Leetcode: Reorder List

FROM

思路

1. 将后半段截取下来再倒序, 插入到前半段, 时间复杂度为 o(n)

代码

#include <iostream>
using namespace std;struct ListNode {int val;ListNode *next;ListNode(int x) : val(x), next(NULL) {}};class Solution {
public:void reorderList(ListNode *head) {if(head == NULL)return;int size = 0;ListNode *curNode = head;while(curNode) {size++;curNode = curNode->next;}    int step = size - (size >> 1); // pick the larger partListNode *cursor1 = head, *cursor2 = head;while(--step)cursor2 = cursor2->next;ListNode *tmp = cursor2;if(tmp != NULL) {cursor2 = tmp->next;tmp->next = NULL;}cursor2 = reverseList(cursor2);while(cursor2) {ListNode *tmp = cursor2;cursor2 = cursor2->next;tmp->next = cursor1->next;cursor1->next = tmp;cursor1 = tmp->next;}}ListNode* reverseList(ListNode* head) {if(head == NULL)return head;ListNode *curNode = head;while(curNode->next) {ListNode *nextNode = curNode->next;curNode->next = nextNode->next;nextNode->next = head;head = nextNode;}return head;}
};int main() {int arr[10] = {1, 2, 3, 4, 5, 6, 7, 8};ListNode* list[10];int len = 3;for(int i = 0; i < len; i ++) list[i] = new ListNode(arr[i]);for(int i = 0; i < len-1; i ++)list[i]->next = list[i+1];(new Solution())->reorderList(list[0]);ListNode *lists = list[0];while(lists) {cout << lists->val << " ";lists = lists->next;}cout << endl;return 0;
}

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