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.
/*
* @lc app=leetcode id=503 lang=cpp
*
* [503] Next Greater Element II
*
* https://leetcode.com/problems/next-greater-element-ii/description/
*
* algorithms
* Medium (51.85%)
* Likes: 791
* Dislikes: 48
* Total Accepted: 57.8K
* Total Submissions: 111.2K
* Testcase Example: '[1,2,1]'
*
*
* 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.
*
*/
class Solution {
public:
vector<int> nextGreaterElements(vector<int>& nums) {
const int N = nums.size();
vector<int> res(N, -1);
stack<int> s;
for (int i = 0; i < nums.size() * 2; ++i) {
const int num = nums[i % N];
while (!s.empty() && nums[s.top()] < num) {
res[s.top()] = num;
s.pop();
}
if (i < N) s.push(i);
}
return res;
}
};