LeetCode503. Next Greater Element II

题目：

Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element. The Next Greater Number of a number x is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number. If it doesn't exist, output -1 for this number.

Example 1:

Input: [1,2,1]Output: [2,-1,2]Explanation: The first 1's next greater number is 2; 
The number 2 can't find next greater number; 
The second 1's next greater number needs to search circularly, which is also 2.

Note: The length of given array won't exceed 10000.

解题分析：

这个和Next Greater Element I差不多类似，但是不一样的是，这次只有一个数组，而且后面找不到更大的数，可以从头再找一次，而且这道题会有重复的数出现。因此在上一题，我们用的map的方法就不能用在这道题上了，于是这道题，我们用vector去替代map，例如，在上一题中，map[i] = j（i表示i数后面第一个比它大的数为j），而vector则是用的坐标表示， vector[i] = j）（i表示坐标为i的nums[i]后面第一个比它大的数为nums[j]），一样是用stack的方法去实现。

代码实现：

class Solution {public:vector<int> nextGreaterElements(vector<int>& nums) {stack<int> sta;vector<int> result(nums.size(),-1);for (int i = 0; i < nums.size(); i++) {while (!sta.empty() && nums[sta.top()] < nums[i]) {result[sta.top()] = nums[i];sta.pop();}sta.push(i);}for (int i = 0; i < nums.size(); i++) {while (!sta.empty() && nums[sta.top()] < nums[i]) {result[sta.top()] = nums[i];sta.pop();}}return result;}};