• 链表操作算法题合集


    0.单链表的增、删、改、查(无头指针)

      1 #include <stdio.h>
      2 #include<stdlib.h>
      3 struct Node
      4 {
      5     int val;
      6     Node * next;
      7 };
      8 
      9 Node* Node_Insert(Node* First,int val) 
     10 {
     11     Node* p=(Node*)calloc(1,sizeof(Node));
     12     p->val=val;
     13     p->next=First;
     14     First=p;
     15     return First;
     16 }
     17 
     18 Node* Node_Delete(Node* First,int val) 
     19 {
     20     Node* p = First;
     21     Node* pre=NULL;
     22     Node* tem=NULL;
     23     int find=0;
     24     while(p!=NULL)
     25     {
     26         if(p->val==val)
     27         {
     28             pre->next=p->next;
     29             tem=p;
     30             p=pre->next;
     31             free(tem);
     32         }
     33         else
     34         {
     35             pre=p;
     36             p=p->next;
     37         }
     38     }
     39     return First;
     40 }
     41 
     42 
     43 
     44 Node* Node_Change(Node* First,int m,int n)
     45 {
     46     Node* p = First;
     47     while(p!=NULL)
     48     {
     49         if(p->val==m)
     50         {
     51             p->val=n;
     52         }
     53         p=p->next;
     54     }
     55     return First;
     56 }
     57 
     58 
     59 int Node_Search(Node* First,int x)
     60 {
     61     Node* p = First;
     62     int index=0;
     63     while(p!=NULL)
     64     {
     65         ++index;
     66         if(p->val==x)
     67             break;
     68         p=p->next;
     69     }
     70     if(p==NULL)
     71         index=-1;
     72     return index;
     73 }
     74 
     75 
     76 void order(Node* First)
     77 {
     78     Node* p=First;
     79     while(p!=NULL)
     80     {
     81         printf("%d ",p->val);
     82         p=p->next;
     83     }
     84 }
     85 
     86 
     87 int main()
     88 {
     89     int n,i,tem,tem2;
     90     while(printf("链表长度:")&&scanf("%d",&n)!=EOF)
     91     {
     92         getchar();
     93         Node* First=NULL;
     94         printf("请插入链表:");
     95         for(i=0;i<n;i++)
     96         {
     97             scanf("%d",&tem);
     98             First=Node_Insert(First,tem);
     99         }
    100         getchar();
    101         printf("请输入删除的值:");
    102         scanf("%d",&tem);
    103         getchar();
    104         First=Node_Delete(First,tem);
    105         printf("删除后:
    ");
    106         order(First);
    107         printf("
    ");
    108         printf("请输入修改前和修改后的值:");
    109         scanf("%d %d",&tem,&tem2);
    110         getchar();
    111         First=Node_Change(First,tem,tem2);
    112         printf("修改后:
    ");
    113         order(First);
    114         printf("
    ");
    115         printf("请输入要查找的值:");
    116         scanf("%d",&tem);
    117         getchar();
    118         tem2=Node_Search(First,tem);
    119         if(tem2==-1) printf("没有这个数
    ");
    120         else printf("第%d个数
    ",tem2);
    121     }
    122     return 0;
    123 }

     

    1.单链表反转

     1 /*
     2 struct ListNode {
     3     int val;
     4     struct ListNode *next;
     5     ListNode(int x) :
     6             val(x), next(NULL) {
     7     }
     8 };*/
     9 class Solution {
    10 public:
    11     ListNode* ReverseList(ListNode* pHead) {
    12           if(pHead!=NULL)
    13          {
    14                 ListNode * pre=NULL;
    15                 ListNode * p=pHead;
    16                ListNode * q=pHead->next;
    17             while(q!=NULL)
    18             {
    19                 p->next=pre;
    20                 pre=p;
    21                 p=q;
    22                 q=q->next;
    23             }
    24             p->next=pre;
    25             return p;
    26          }
    27          return NULL;
    28     }
    29 };

    2.找出单链表的倒数第n个元素

    *p,*q 指向第一个指针,p向前移动n次,q再跟着和p一直移动,等p移到末尾,q所指的就是倒是第n个元素

     1 #include <stdio.h>
     2 #include<stdlib.h>
     3 struct Node
     4 {
     5     int val;
     6     Node * next;
     7 };
     8 
     9 Node* Node_Insert(Node* First,int val) 
    10 {
    11     Node* p=(Node*)calloc(1,sizeof(Node));
    12     p->val=val;
    13     p->next=First;
    14     First=p;
    15     return First;
    16 }
    17 
    18 
    19 
    20 
    21 void order(Node* First)
    22 {
    23     Node* p=First;
    24     while(p!=NULL)
    25     {
    26         printf("%d ",p->val);
    27         p=p->next;
    28     }
    29 }
    30 
    31 int My_Node_Search(Node* First,int n)
    32 {
    33     int i;
    34     Node* p,*q;
    35     q=p=First;
    36 
    37     if(First->next==NULL&&n==1) return 1;
    38     if(First->next==NULL&&n>1) return -1;
    39     if(n<=0) return -1;
    40 
    41     for(i=0;i<n;i++)
    42     {
    43         if(p==NULL) return -1;
    44         p=p->next;
    45     }
    46     while(p!=NULL)
    47     {
    48         p=p->next;
    49         q=q->next;
    50     }
    51     return q->val;
    52 }
    53 
    54 int main()
    55 {
    56     int n,i,tem;
    57     while(printf("链表长度:")&&scanf("%d",&n)!=EOF)
    58     {
    59         getchar();
    60         Node* First=NULL;
    61         printf("请插入链表:");
    62         for(i=0;i<n;i++)
    63         {
    64             scanf("%d",&tem);
    65             First=Node_Insert(First,tem);
    66         }
    67         getchar();
    68         printf("链表为:");
    69         order(First);
    70         printf("
    ");
    71 
    72         printf("取倒数第几个数:");
    73         while(scanf("%d",&tem)!=EOF)
    74         {
    75             tem=My_Node_Search(First,tem);
    76             if(tem!=-1)
    77                 printf("值为%d
    ",tem);
    78             else printf("越界!
    ");
    79             printf("取倒数第几个数:");
    80         }
    81     }
    82     return 0;
    83 }

    3.找出单链表的中间元素(见0)


    4.删除无头单链表的一个节点(见0)


    5.两个不交叉的有序链表的合并

     1 /*
     2 struct ListNode {
     3     int val;
     4     struct ListNode *next;
     5     ListNode(int x) :
     6             val(x), next(NULL) {
     7     }
     8 };*/
     9 class Solution {
    10 public:
    11     ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
    12     {
    13          ListNode* p=pHead1;
    14         ListNode* q=pHead2;
    15             ListNode* First=NULL;
    16             ListNode* r=NULL;
    17         if (p == NULL )
    18             return q;
    19         if (q == NULL )
    20             return p;
    21         
    22         if(p->val<q->val)
    23         {
    24             r=p;
    25             p=p->next;
    26         }
    27         else
    28         {
    29             r=q;
    30             q=q->next;
    31         }
    32 
    33     First=r;
    34     while(p!=NULL&&q!=NULL)
    35     {
    36         if(p->val<q->val)
    37         {
    38             r->next=p;
    39             p=p->next;
    40         }
    41         else
    42         {
    43             r->next=q;
    44             q=q->next;
    45         }
    46         r=r->next;
    47     }
    48 
    49     while(p!=NULL)
    50     {
    51         r->next=p;
    52         r=r->next;
    53         p=p->next;
    54     }
    55 
    56     while(q!=NULL)
    57     {
    58         r->next=q;
    59         r=r->next;
    60         q=q->next;
    61     }
    62 
    63     r=NULL;
    64 
    65     return First;
    66     }
    67 };

    6.判断单链表是否有环?

    p1 p2 指向开头,p1的每次移动步长为1,p2的每次移动的步长为2。在p2==NULL前,如果出现p1==p2,则有环。
     1 #include <stdio.h>
     2 #include<stdlib.h>
     3 struct Node
     4 {
     5     int val;
     6     Node * next;
     7 };
     8 
     9 Node* Node_Create(Node* p,int val) 
    10 {
    11     p=(Node*)calloc(1,sizeof(Node));
    12     p->val=val;
    13     p->next=NULL;
    14     return p;
    15 }
    16 
    17 int If_Have_Circle(Node* First)
    18 {
    19     if( First==NULL) return 0;
    20     Node* p1=First;
    21     Node* p2=First;
    22     int Have=0;
    23     while(p1!=NULL&&p2!=NULL)
    24     {
    25         p1=p1->next;
    26         if(p1==NULL) break;
    27 
    28         p2=p2->next;
    29         if(p2==NULL) break;
    30         p2=p2->next;
    31         if(p2==NULL) break;
    32 
    33         if(p1==p2)
    34         {
    35             Have=1;
    36             break;
    37         }
    38     }
    39 
    40     return Have;
    41 }
    42 int main()
    43 {
    44     int i,tem;
    45 
    46     Node* No_Circle_First=NULL;
    47     Node* Have_Circle_First=NULL;
    48 
    49     Node* p=NULL;
    50     Node* q=NULL;
    51     q=Node_Create(q,1);
    52     p=q;
    53     No_Circle_First=q;
    54 
    55     q=Node_Create(q,2);
    56     p->next=q;
    57     p=p->next;
    58     q=Node_Create(q,3);
    59     p->next=q;
    60     p=p->next;
    61     q=Node_Create(q,4);
    62     p->next=q;
    63 
    64     q=Node_Create(q,1);
    65     p=q;
    66     Have_Circle_First=q;
    67 
    68     q=Node_Create(q,2);
    69     p->next=q;
    70     p=p->next;
    71 
    72     q=Node_Create(q,3);
    73     p->next=q;
    74     p=p->next;
    75     Node* index=p;
    76 
    77     q=Node_Create(q,4);
    78     p->next=q;
    79     p=p->next;
    80 
    81     q=Node_Create(q,5);
    82     p->next=q;
    83     p=p->next;
    84     p->next=index;
    85 
    86     int tem1=If_Have_Circle(No_Circle_First);
    87     printf("%s
    ",tem1==0?"No":"Yes");
    88     tem1=If_Have_Circle(Have_Circle_First);
    89     printf("%s
    ",tem1==0?"No":"Yes");
    90     return 0;
    91 }

    7.判断两个单链表是否相交(见8)


    8.两个单链表相交,计算相交点

    先计算出两条链表的长度L1、L2,n=ans(L1-L2)。p1,p2分别指向两个链表的开头,指向长的链表的指针先向前移动n次,然后再两个一起移动。如果出现p1==p2,则开始计数。
     1 #include <stdio.h>
     2 #include<stdlib.h>
     3 struct Node
     4 {
     5     int val;
     6     Node * next;
     7 };
     8 
     9 Node* Node_Create(Node* p,int val) 
    10 {
    11     p=(Node*)calloc(1,sizeof(Node));
    12     p->val=val;
    13     p->next=NULL;
    14     return p;
    15 }
    16 
    17 
    18 void Find_Same(Node* First1,Node* First2)
    19 {
    20     int len1,len2,count;
    21     len1=len2=count=0;
    22     Node* p,*q;
    23     p=First1;
    24     while(p!=NULL)
    25     {
    26         ++len1;
    27         p=p->next;
    28     }
    29 
    30     p=First2;
    31     while(p!=NULL)
    32     {
    33         ++len2;
    34         p=p->next;
    35     }
    36     p=First1;q=First2;
    37     while(len1!=len2)
    38     {
    39         if(len1>len2) 
    40         {
    41             p=p->next;
    42             ++len2;
    43         }
    44         else
    45         {
    46             q=q->next;
    47             ++len1;
    48         }
    49     }
    50 
    51     while(p!=q)
    52     {
    53         p=p->next;
    54         q=q->next;
    55     }
    56 
    57     while(p!=NULL)
    58     {
    59         ++count;
    60         printf("%d ",p->val);
    61         p=p->next;
    62     }
    63     printf("
    共%d个公共节点
    ",count);
    64 }
    65 
    66 int main()
    67 {
    68     int i;
    69 
    70     Node* First1=NULL;
    71     Node* First2=NULL;
    72 
    73     Node* p=NULL;
    74     Node* q=NULL;
    75     q=Node_Create(q,1);
    76     p=q;
    77     First1=q;
    78 
    79     q=Node_Create(q,2);
    80     p->next=q;
    81     p=p->next;
    82     q=Node_Create(q,3);
    83     p->next=q;
    84     p=p->next;
    85     Node* index=p;
    86     q=Node_Create(q,4);
    87     p->next=q;
    88 
    89     q=Node_Create(q,1);
    90     p=q;
    91     First2=q;
    92     p->next=index;
    93     Find_Same(First1,First2);
    94     return 0;
    95 }

    9.单链表排序

    链表只能相邻改变,所以用冒泡排序最好。
     1 #include <stdio.h>
     2 #include<stdlib.h>
     3 struct Node
     4 {
     5     int val;
     6     Node * next;
     7 };
     8 
     9 Node* Node_Insert(Node* First,int val) 
    10 {
    11     Node* p=(Node*)calloc(1,sizeof(Node));
    12     p->val=val;
    13     p->next=First;
    14     First=p;
    15     return First;
    16 }
    17 
    18 void order(Node* First)
    19 {
    20     Node* p=First;
    21     while(p!=NULL)
    22     {
    23         printf("%d ",p->val);
    24         p=p->next;
    25     }
    26 }
    27 
    28 
    29 Node* Node_sort(Node* First)
    30 {
    31     if(First==NULL)
    32         return First;
    33     Node* p1,*p2,*pre;
    34     Node* rear=First;
    35     Node* L=(Node*)calloc(1,sizeof(Node));
    36     L->next=First;
    37     while(rear->next!=NULL)
    38         rear=rear->next;
    39     while(rear!=First)
    40     {
    41         pre=L;
    42         p1=L->next;
    43         p2=L->next->next;
    44         while(p2!=rear->next&&p2!=NULL)
    45         {
    46             if(p1->val>p2->val)
    47             {
    48                 pre->next=p1->next;
    49                 p1->next=p2->next;
    50                 p2->next=p1;
    51                 pre=pre->next;
    52                 p2=p1->next;
    53             }
    54             else
    55             {
    56                 pre=pre->next;
    57                 p1=p1->next;
    58                 p2=p2->next;
    59             }
    60         }
    61 
    62         rear=pre;
    63     }
    64 
    65     return L->next;
    66 }
    67 
    68 int main()
    69 {
    70     int n,i,tem;
    71     while(printf("链表长度:")&&scanf("%d",&n)!=EOF)
    72     {
    73         getchar();
    74         Node* First=NULL;
    75         printf("请插入链表:");
    76         for(i=0;i<n;i++)
    77         {
    78             scanf("%d",&tem);
    79             First=Node_Insert(First,tem);
    80         }
    81         getchar();
    82         printf("链表为:");
    83         order(First);
    84         printf("
    ");
    85 
    86         First=Node_sort(First);
    87         printf("排序后为:");
    88         order(First);
    89         printf("
    ");
    90     }
    91     return 0;
    92 }

    10.删除单链表中重复的元素

    用Hash数组保存各相同元素值的个数
     1 #include <stdio.h>
     2 #include<stdlib.h>
     3 
     4 struct Node
     5 {
     6     int val;
     7     Node * next;
     8 };
     9 
    10 Node* Node_Insert(Node* First,int val) 
    11 {
    12     Node* p=(Node*)calloc(1,sizeof(Node));
    13     p->val=val;
    14     p->next=First;
    15     First=p;
    16     return First;
    17 }
    18 
    19 void order(Node* First)
    20 {
    21     Node* p=First;
    22     while(p!=NULL)
    23     {
    24         printf("%d ",p->val);
    25         p=p->next;
    26     }
    27 }
    28 
    29 int cnt[101];
    30 
    31 Node* Delete_Same(Node* First)
    32 {
    33     Node* p=First;
    34     Node* pre,*r;
    35     while(p!=NULL)
    36     {
    37         ++cnt[p->val];
    38         p=p->next;
    39     }
    40     Node* L=(Node*)calloc(1,sizeof(Node));
    41     L->next=First;
    42     pre=L;
    43     p=First;
    44     while(p!=NULL)
    45     {
    46         if(cnt[p->val]>1)
    47         {
    48             --cnt[p->val];
    49             r=p;
    50             p=p->next;
    51             pre->next=p;
    52             free(r);
    53         }
    54         else
    55         {
    56             p=p->next;
    57             pre=pre->next;
    58         }
    59     }
    60 
    61     return L->next;
    62 }
    63 
    64 int main()
    65 {
    66     int n,i,tem,tem2;
    67     while(printf("链表长度:")&&scanf("%d",&n)!=EOF)
    68     {
    69         getchar();
    70         Node* First=NULL;
    71         printf("请插入链表:");
    72         for(i=0;i<n;i++)
    73         {
    74             scanf("%d",&tem);
    75             First=Node_Insert(First,tem);
    76         }
    77         printf("原链表:");
    78         order(First);
    79         printf("
    ");
    80 
    81         for(i=0;i<101;i++)
    82             cnt[i]=0;
    83         First=Delete_Same(First);
    84 
    85         printf("删除后:");
    86         order(First);
    87         printf("
    ");
    88 
    89     }
    90     return 0;
    91 }


    11 链表拆分(将链表奇数位置上的节点构成一个新链表,偶数位置上的结点构成一个新链表)

     1 #include <stdio.h>
     2 #include<stdlib.h>
     3 struct Node
     4 {
     5     int val;
     6     Node * next;
     7 };
     8 
     9 Node* Node_Insert(Node* First,int val) 
    10 {
    11     Node* p=(Node*)calloc(1,sizeof(Node));
    12     p->val=val;
    13     p->next=First;
    14     First=p;
    15     return First;
    16 }
    17 
    18 
    19 void order(Node* First)
    20 {
    21     Node* p=First;
    22     while(p!=NULL)
    23     {
    24         printf("%d ",p->val);
    25         p=p->next;
    26     }
    27 }
    28 
    29 void Node_Cut(Node* First,Node* &First1,Node* &First2)
    30 {
    31     if(First==NULL) 
    32         return;
    33     if(First->next==NULL)
    34     {
    35         First1=First;
    36         First1->next=NULL;
    37         return;
    38     }
    39 
    40     First1=First;
    41     First2=First->next;
    42 
    43     Node* p1=First1;
    44     Node* p2=First2;
    45     while(p1!=NULL&&p2!=NULL)
    46     {
    47         if(p1->next==NULL) break;
    48         if(p1->next->next==NULL) break;
    49         p1->next=p1->next->next;
    50         p1=p1->next;
    51 
    52         if(p2->next==NULL) break;
    53         if(p2->next->next==NULL) break;
    54         p2->next=p2->next->next;
    55         p2=p2->next;
    56     }
    57     p1->next=NULL;
    58     p2->next=NULL;
    59 }
    60 
    61 int main()
    62 {
    63     int n,i,tem,tem2;
    64     while(printf("链表长度:")&&scanf("%d",&n)!=EOF)
    65     {
    66         getchar();
    67         Node* First=NULL;
    68         Node* First1=NULL;
    69         Node* First2=NULL;
    70         printf("请插入链表:");
    71         for(i=0;i<n;i++)
    72         {
    73             scanf("%d",&tem);
    74             First=Node_Insert(First,tem);
    75         }
    76         printf("原链表:");
    77         order(First);
    78         printf("
    ");
    79 
    80         Node_Cut(First,First1,First2);
    81         printf("拆分后:
    ");
    82         order(First1);
    83         printf("
    ");
    84         order(First2);
    85         printf("
    ");
    86 
    87     }
    88     return 0;
    89 }

    12 大整数加法

      1 #include<stdio.h>
      2 #include<stdlib.h>
      3 
      4 struct Node
      5 {
      6     int val;
      7     Node * next;
      8 };
      9 
     10 Node* Node_Insert(Node* First,int val) 
     11 {
     12     Node* p=(Node*)calloc(1,sizeof(Node));
     13     p->val=val;
     14     p->next=First;
     15     First=p;
     16     return First;
     17 }
     18 
     19 
     20 void order(Node* First)
     21 {
     22     Node* p=First;
     23     while(p!=NULL)
     24     {
     25         printf("%d",p->val);
     26         p=p->next;
     27     }
     28 }
     29 
     30 int jinwei,tuiwei;
     31 
     32 Node* add1(Node* Long_Num,Node* Short_Num)// 同号整数相加
     33 {
     34     Node* p1,*p2;
     35     p1=Long_Num;p2=Short_Num;
     36     while(p2!=NULL)
     37     {
     38         p1->val+=p2->val;
     39         p1=p1->next;
     40         p2=p2->next;
     41     }
     42 
     43     Node* p3=Long_Num;
     44     while(p3!=p1)
     45     {
     46         if(p3->val>=10)
     47         {
     48             if(p3->next==NULL)
     49             {
     50                 jinwei=1;//进位
     51                 p3->val=p3->val%10;
     52             }
     53             else
     54             {
     55                 p3->next->val+=p3->val/10;
     56                 p3->val=p3->val%10;
     57             }
     58         }
     59         p3=p3->next;
     60     }
     61 
     62     return Long_Num;
     63 }
     64 
     65 Node* add2(Node* Long_Num,Node* Short_Num)//异号相加
     66 {
     67     Node* p1,*p2;
     68     p1=Long_Num;p2=Short_Num;
     69     while(p2!=NULL)
     70     {
     71         p1->val-=p2->val;
     72         p1=p1->next;
     73         p2=p2->next;
     74     }
     75     Node* p3=Long_Num;
     76     while(p3!=p1)
     77     {
     78         if(p3->val<0)
     79         {
     80 
     81             --p3->next->val;
     82             p3->val+=10;
     83         }
     84         p3=p3->next;
     85     }
     86     return Long_Num;
     87 }
     88 
     89 int My_Compare(Node* Num1,Node* Num2)
     90 {
     91     Node* p1,*p2;
     92     p1=Num1;
     93     p2=Num2;
     94     while(p1!=NULL)
     95     {
     96         if(p1->val>p2->val) return 1;
     97         if(p1->val<p2->val) return -1;
     98         p1=p1->next;
     99         p2=p2->next;
    100     }
    101     return 0;
    102 }
    103 
    104 Node* Chain_Reverse(Node* First)
    105 {
    106     if(First!=NULL)
    107     {
    108         Node * pre=NULL;
    109         Node * p=First;
    110         Node * q=First->next;
    111         while(q!=NULL)
    112         {
    113             p->next=pre;
    114             pre=p;
    115             p=q;
    116             q=q->next;
    117         }
    118         p->next=pre;
    119         return p;
    120     }
    121     return First;
    122 
    123 }
    124 
    125 int main()
    126 {
    127     printf("输入格式为:
    Num1
    +
    Num2
    ");
    128     while(1)
    129     {
    130         Node* Num1,* Num2;
    131         Num1=NULL;
    132         Num2=NULL;
    133         char tem;
    134         int fir=1;
    135         int fu1,fu2;
    136         int len1,len2;
    137         char Highest_Val1,Highest_Val2;
    138         len1=len2=0;
    139         while(scanf("%c",&tem))//第一个数
    140         {
    141             if(tem=='
    ') break;
    142             if(fir)
    143             {
    144                 fir=0;
    145                 if(tem=='-')
    146                 {
    147                     fu1=1;
    148                     scanf("%c",&Highest_Val1);
    149                     Num1=Node_Insert(Num1,Highest_Val1-'0');
    150                     ++len1;
    151                 }
    152                 else
    153                 {
    154                     fu1=0;
    155                     ++len1;
    156                     Highest_Val1=tem;
    157                     Num1=Node_Insert(Num1,tem-'0');
    158                 }
    159 
    160             }
    161             else 
    162             {   ++len1;
    163             Num1=Node_Insert(Num1,tem-'0');
    164             }
    165 
    166         }
    167         getchar();
    168         getchar();
    169 
    170         fir=1;
    171         while(scanf("%c",&tem))//第二个数
    172         {
    173             if(tem=='
    ') break;
    174             if(fir)
    175             {
    176                 fir=0;
    177                 if(tem=='-') 
    178                 {
    179                     fu2=1;
    180                     scanf("%c",&Highest_Val2);
    181                     Num2=Node_Insert(Num2,Highest_Val2-'0');
    182                     ++len2;
    183                 }
    184                 else
    185                 {
    186                     fu2=0;
    187                     ++len2;
    188                     Highest_Val2=tem;
    189                     Num2=Node_Insert(Num2,tem-'0');
    190                 }
    191             }
    192             else
    193             {
    194                 ++len2;
    195                 Num2=Node_Insert(Num2,tem-'0');    
    196             }
    197         }
    198         Node* result;
    199         jinwei=tuiwei=0;
    200         if(fu1==0&&fu2==0) //全正
    201         {
    202             if(len1>len2)  result=add1(Num1,Num2);
    203             else  result=add1(Num2,Num1);
    204             result=Chain_Reverse(result);
    205             if(jinwei) printf("1");
    206             order(result);
    207             printf("
    ");
    208         }
    209         else if(fu1==1&&fu2==1) //全负
    210         {
    211             if(len1>len2)  result=add1(Num1,Num2);
    212             else  result=add1(Num2,Num1);
    213             result=Chain_Reverse(result);
    214             printf("-");
    215             if(jinwei) printf("1");
    216             order(result);
    217             printf("
    ");
    218         }
    219         else if((fu1==1&&fu2==0)||(fu1==0&&fu2==1)) //正负相加
    220         {
    221             int bigger=1;
    222             int cmp=100;
    223             if(len1>len2)  result=add2(Num1,Num2);
    224             else if(len1<len2) 
    225             {
    226                 bigger=2;
    227                 result=add2(Num2,Num1);
    228             }
    229             else if(len1==len2)
    230             {
    231                 Num1=Chain_Reverse(Num1);
    232                 Num2=Chain_Reverse(Num2);
    233                 cmp=My_Compare(Num1,Num2);
    234                 Num1=Chain_Reverse(Num1);
    235                 Num2=Chain_Reverse(Num2);
    236                 if(cmp==1) result=add2(Num1,Num2);
    237                 else if(cmp==-1) 
    238                 {
    239                     bigger=2;
    240                     result=add2(Num2,Num1);
    241                 }
    242             }
    243 
    244             if(cmp==0) printf("0
    ");
    245             else
    246             {
    247                 result=Chain_Reverse(result);
    248                 Node* p=result;
    249                 while(p->val==0)
    250                     p=p->next;
    251                 if(bigger==1)
    252                 {
    253                     if(fu1) printf("-");
    254                 }
    255                 else if(bigger==2)
    256                 {
    257                     if(fu2)   printf("-");
    258                 }
    259                 while(p!=NULL)
    260                 {
    261                     printf("%d",p->val);
    262                     p=p->next;
    263                 }
    264                 printf("
    ");
    265             }
    266         }
    267 
    268     }
    269     return 0;
    270 }
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  • 原文地址:https://www.cnblogs.com/xiaoyesoso/p/4234443.html
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