Leetcode 503. Next Greater Element II (单调栈好题)

503. Next Greater Element II
     Medium
     Given a circular integer array nums (i.e., the next element of nums[nums.length - 1] is nums[0]), return the next greater number for every element in nums.

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, return -1 for this number.

Example 1:

Input: nums = [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.
Example 2:

Input: nums = [1,2,3,4,3]
Output: [2,3,4,-1,4]

Constraints:

1 <= nums.length <= 104
-109 <= nums[i] <= 109

解法1：单调栈

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