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.
Related Topics:
Stack
Similar Questions:
// OJ: https://leetcode.com/problems/next-greater-element-ii/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
vector<int> nextGreaterElements(vector<int>& A) {
stack<int> s;
int N = A.size();
vector<int> ans(N, -1);
for (int i = 0; i < 2 * N; ++i) {
int n = A[i % N];
while (s.size() && A[s.top()] < n) {
ans[s.top()] = n;
s.pop();
}
s.push(i % N);
}
return ans;
}
};