题目来源LeetCode
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.
解题思路：将数组复制两遍来进行circular，然后再进行比较找出下一个最大元素，具体代码如下：
class Solution {public: vector<int> nextGreaterElements(vector<int>& nums) { vector<int>double_nums; vector<int>results; int l = nums.size(); for(int i = 0; i < l; i++){ double_nums.push_back(nums[i]); } for(int i = 0; i < l; i++){ double_nums.push_back(nums[i]); } for(int i = 0; i < l; i++){ for(int j = i; j <= i+l; j++){ if(double_nums[i] < double_nums[j]){ results.push_back(double_nums[j]); break; } else if(j == i+l && double_nums[i] >= double_nums[j]){ results.push_back(-1); break; } } } return results; }};新闻热点
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