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。。。