【Leetcode】【Easy】Merge Two Sorted Lists

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

解题：

普通做法三步：

1、常规边界判定

2、由于最终要返回合成的链表头部，因此需要先比较l1和l2首个结点值大小，确定ret值；

3、新建一个cur指针操作要返回的链表，串起l1和l2；

记录：

1、指针赋值的关注点在等号的右边。指针名是变量名，随着赋值操作随时改变，不会产生连带影响；

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
        ListNode *ret = NULL;
        ListNode *cur = NULL;
        
        if (!l1) return l2;
        if (!l2) return l1;
            
        if (l1->val <= l2->val) {
            ret = l1;
            l1 = l1->next;
        } else { 
            ret = l2;
            l2 = l2->next;
        }
        
        cur = ret;
        while (l1 && l2) {
            if (l1->val <= l2->val) {
                cur->next = l1;
                l1 = l1->next;
            } else {
                cur->next = l2;
                l2 = l2->next;
            }
            cur = cur->next;
        }
        
        if (!l1) 
            cur->next = l2;
        else 
            cur->next = l1;
        
        return ret;
    }
};

文艺做法：

使用二重指针解决了判定链表头大小的步骤

类似另一道简单题Remove Nth Node From End of List

Linus once said " People who understand pointers just use a“pointer to the entry pointer”"

class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode* head = NULL, **t = &head;
        while(l1 != NULL && l2 != NULL) {
            if(l1->val < l2->val) {
                *t = l1;
                l1 = l1->next;
            } else {
                *t = l2;
                l2 = l2->next;
            }
            t = &((*t)->next);
        }
        *t = (l1 != NULL ? l1 : l2);
        return head;
    }
};