Leetcode ☞ 143. Reorder List

143. Reorder List

My Submissions

Question

Total Accepted: 60546 Total Submissions: 270201 Difficulty: Medium

Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You must do this in-place without altering the nodes' values.

For example,
Given {1,2,3,4}, reorder it to {1,4,2,3}.

我的AC：（16ms，最快一批）

struct ListNode* reverseList(struct ListNode* head){    if(!head || !head->next) return head;    struct ListNode *p, *p1, *p2;    p2 = head;    p1 = NULL;    while(p2){        p = p2->next;        p2->next = p1;        p1 = p2;        p2 = p;    }    return p1;}void reorderList(struct ListNode* head) {    if(!head || !head->next) return;    struct ListNode *fast, *slow, *secondPartHead;    fast = slow = head;        while(fast->next && fast->next->next){        fast = fast->next->next;        slow = slow->next;    }    secondPartHead = slow->next;    slow->next = NULL;    secondPartHead = reverseList(secondPartHead);        struct ListNode *firstcurr = head;    struct ListNode *secondcurr = secondPartHead;    struct ListNode *p1, *p2;    while(firstcurr && secondcurr){        p1 = firstcurr->next;        p2 = secondcurr->next;        firstcurr->next = secondcurr;        secondcurr->next = p1;        firstcurr = p1;        secondcurr = p2;    }}

思路：

快慢指针找中位数-->剪掉后半部-->反转后半部-->合并

注意：

1、reorderList函数类型是void。

2、最后的循环条件while(firstcurr && secondcurr【一开始落了】)。

例子：

【1,2,3,4,5,6】-->【1,2,3】和【6,5,4】-->【1,6,2,5,3,null】，【4,null】。最后一步：把4插到了3跟null中间，firstcurr 和secondcurr都是NULL。

【1,2,3,4,5,6,7】-->【1,2,3,4】和【7,6,5】-->【1,7,2,6,3,4,null】，【5,null】。最后一步：把5插到了3跟4中间，secondcurr为NULL，而firstcurr不是NULL！

所以循环条件写成while(secondcurr)即可。