【LeetCode】19

Given a linked list, remove the nth node from the end of list and return its head.

For example,

   Given linked list: 1->2->3->4->5, and n = 2.

   After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:
Given n will always be valid.
Try to do this in one pass.

Solution 1:  重点在于one pass，以空间换取时间，使用vector<ListNode*>记录list的节点，每个指针指向一个链表节点，遍历一遍list算出size，通过下标计算（size-n）即可定位到需要删除的节点

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {　　//runtime:12ms
        vector<ListNode*> vec;
        ListNode *temp=head;
        int size=0;
        while(temp){
            vec.push_back(temp);
            temp=temp->next;
            size++;
        }
        vec.push_back(NULL);//必不可少，否则如果删除的是倒数第一个元素的话下面的vec[i-1]->next=vec[i+1]中i+1溢出
        int i=size-n;
        if(i==0)//delete head
            return head->next;
        else{
            vec[i-1]->next=vec[i+1];
            return head;
        }
    }
};

Solution 2: 不需要额外的空间，当正数第n个移到链表尾时，head移到倒数第n个

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {　　　　//runtime:4ms
        ListNode *temp=head;
        while(--n)temp=temp->next;
        ListNode *del=head,*predel=NULL;
        while(temp->next){
            temp=temp->next;
            predel=del;
            del=del->next;
        }
        if(predel==NULL)return head->next;
        else{
            predel->next=del->next;
            return head;
        }
        
    }
};