Merge Two Sorted Lists Leetcode

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.

思想和merge sort一样，只是要考虑相等的情况的时候该怎么办。

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode(int x) { val = x; }* }*/
public class Solution {public ListNode mergeTwoLists(ListNode l1, ListNode l2) {if (l1 == null) {return l2;}if (l2 == null) {return l1;}ListNode result = new ListNode(0);ListNode returnResult = result;while (l1 != null && l2 != null) {while (l1 != null && l2 != null && l1.val < l2.val) {result.next = l1;l1 = l1.next;result = result.next;}while (l1 != null && l2 != null && l1.val > l2.val) {result.next = l2;l2 = l2.next;result = result.next;}while (l1 != null && l2 != null && l1.val == l2.val) {result.next = l1;l1 = l1.next;result = result.next;result.next = l2;l2 = l2.next;result = result.next;}}if (l1 == null) {result.next = l2;} else {result.next = l1;}return returnResult.next;}
}

写的还是有些冗余了。。。

简洁版：

public class Solution {public ListNode mergeTwoLists(ListNode l1, ListNode l2) {if (l1 == null) {return l2;}if (l2 == null) {return l1;}ListNode result = new ListNode(0);ListNode returnResult = result;while (l1 != null && l2 != null) {if (l1.val < l2.val) {result.next = l1;l1 = l1.next;} else {result.next = l2;l2 = l2.next;}result = result.next;}if (l1 == null) {result.next = l2;} else {result.next = l1;}return returnResult.next;}
}

没想到还可以用recursion做

public class Solution {public ListNode mergeTwoLists(ListNode l1, ListNode l2) {if (l1 == null) {return l2;}if (l2 == null) {return l1;}ListNode result = null;if (l1.val < l2.val) {result = l1;result.next = mergeTwoLists(l1.next, l2);} else {result = l2;result.next = mergeTwoLists(l1, l2.next);}return result;}
}

时隔两个月再来做这道题，发现自己还是进步了的~至少直接写出了简洁版。吼吼吼

也又写了一遍recursion

public class Solution {public ListNode mergeTwoLists(ListNode l1, ListNode l2) {if (l1 == null) {return l2;}if (l2 == null) {return l1;}if (l1.val > l2.val) {l2.next = mergeTwoLists(l1, l2.next);return l2;} else {l1.next = mergeTwoLists(l1.next, l2);return l1;}}
}

C++版本：

/*** 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) {return l2;}if (l2 == NULL) {return l1;}if (l1 -> val > l2 -> val) {l2 -> next = mergeTwoLists(l1, l2 -> next);return l2;} else {l1 -> next = mergeTwoLists(l1 -> next, l2);return l1;}}
};

c++另一种写法怎么也写不过。。。不懂指针什么的，到时候懂了再来写吧。。。#未完待续#需回顾。。。

转载于:https://www.cnblogs.com/aprilyang/p/6620460.html

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