LeetCode：21. Merge Two Sorted Lists

一、問題描述

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.

Example:

Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

二、解題思路

遍歷兩個鏈表，然後比較大小。小的加上，然後繼續比較。直到其中一個到達尾部，那麼就把剩下的鏈表直接加上。

三、代碼實現

/**
 * 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) {
        if(l1 == NULL && l2 == NULL) return NULL;
        else if(l1 != NULL && l2 == NULL) return l1;
        else if(l1 == NULL && l2 != NULL) return l2;
        else {
            ListNode* preHead = new ListNode(0);
            //preHead 需要移動，因此使用resultHead來先保存下
            ListNode* resultHead = preHead;
            while( l1!=NULL && l2!=NULL){
                //同時不爲空，那就比較。然後將小的添加到preHead的尾部
                if(l1->val < l2->val){
                    preHead->next = l1;
                    preHead = preHead->next;
                    l1 = l1->next;
                }else{
                    preHead->next = l2;
                    preHead = preHead->next;
                    l2 = l2->next;
                }
                //while結束，說明有一個爲空了，那就把不爲空的直接加上
                if(l1!=NULL) preHead->next = l1;
                if(l2!=NULL) preHead->next = l2;
            }
            return resultHead->next;
        }
    }
};