Leetcode_sort-list(c++ version)

地址：http://oj.leetcode.com/problems/sort-list/

Sort a linked list in O(n log n) time using constant space complexity.

思路：归并排序，用递归思想。

参考代码：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:ListNode *sortList(ListNode *head) {if(!head || !head->next)return head;ListNode*p1 = head, *p2 = head, *ph = NULL;while(p1){p1 = p1->next;if(p1)p1 = p1->next;ph = p2;p2 = p2->next;}ph->next = NULL;head = sortList(head);p2 = sortList(p2);ListNode*pans = NULL;ListNode**ppcur = &pans;while(head && p2){if(head->val < p2->val){*ppcur = head;head = head->next;}else{*ppcur = p2;p2 = p2->next;}ppcur = &((*ppcur)->next);}if(head)*ppcur = head;else if(p2)*ppcur = p2;return pans;}};

下面的解法可能更好理解啊，先把链表切的短的不能再短(1个node或者0个)，再把这些短的不能再短的链表缝起来。

class Solution {private:ListNode* merge(ListNode*l1, ListNode*l2) {if(!l1)return l2;else if(!l2)return l1;ListNode*head = NULL, *p1 = l1, *p2 = l2, *cur = NULL;while(p1 && p2) {if(p1->val <= p2->val) {if(!head)head = cur = p1;elsecur = cur->next = p1;p1 = p1->next;} else {if(!head)head = cur = p2;elsecur = cur->next = p2;p2 = p2->next;}}if(p1)cur->next = p1;elsecur->next = p2;return head;}void split(ListNode*head, ListNode**l1, ListNode**l2) {ListNode* lnode = *l1, *rnode = *l2;if(!head)lnode = rnode = NULL;else if(!head->next) {lnode = head;rnode = NULL;} else {lnode = rnode = head;ListNode *tmp = NULL;while(lnode) {lnode = lnode->next;if(lnode) {lnode = lnode->next;tmp = rnode;rnode = rnode->next;}}tmp->next = NULL;*l1 = lnode = head;*l2 = rnode;}}public:ListNode *sortList(ListNode *head) {if(!head || !head->next)return head;ListNode*l1 = NULL, *l2 = NULL;split(head, &l1, &l2);l1 = sortList(l1);l2 = sortList(l2);return merge(l1, l2);}};