503. Next Greater Element II

Medium

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.

 1 class Solution {
 2 public:
 3     vector<int> nextGreaterElements(vector<int>& nums) {
 4         vector<int> res(nums.size(),-1);
 5         stack<int> s;
 6         int n=nums.size();
 7         for(int i=0;i<n*2;++i)
 8         {
 9             int num=nums[i%n];
10             while(!s.empty()&&nums[s.top()]<num)
11             {
12                 res[s.top()]=num;
13                 s.pop();
14             }
15             if(i<nums.size())s.push(i);
16         }
17         return res;
18     }
19 };

暴力可以accept, 不过太慢. 空间O(1) 时间 O(n2)

这题使用stack结构是非常完美的, 根据数组的排放顺序, 空间O(1) ~ O(n) 时间O(n)