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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:

Solution 1. Monotonic stack

// 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;
    }
};