[leetcode]503. Next Greater Element II

题目链接：https://leetcode.com/problems/next-greater-element-ii/

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.

方法一（暴力法，时间复杂度O(n*n)）：

class Solution{public:    vector<int> nextGreaterElements(vector<int>& nums)    {        vector<int> res;        int numsSize=nums.size();        for(int i=0;i<numsSize;i++)        {            int tmp=nums[i];            int size=0;            for(int j=i+1;size<numsSize;size++,j++)            {                int k=j%numsSize;                if(nums[k]>tmp)                {                    res.push_back(nums[k]);                    break;                }            }            if(size==numsSize)                res.push_back(-1);        }        return res;    }};

方法二：

解题思路：

初始化一个栈s，s存nums的下标,结果数组res,res值全是-1，对nums遍历两次，如果nums[s.top()]的值小于nums[j],res[s.top()]=nums[j],s出栈，直到nums[s.top()]>=nums[j]，随后s进j。

class Solution{public:    vector<int> nextGreaterElements(vector<int>& nums)    {        vector<int> res(nums.size(),-1);        stack<int> s;        for(int i=0;i<2*nums.size();i++)        {            int j=i%nums.size();            while (!s.empty() && nums[s.top()]<nums[j])            {                res[s.top()]=nums[j];                s.pop();            }            s.push(j);        }        return res;    }};