503. Next Greater Element II

https://leetcode.com/problems/next-greater-element-ii/description/

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.

循环数组问每个数后面第一个比它大的。根据值维持递减栈，直到遇到i比栈顶大的，不断弹出比i小的元素，弹出的元素的右边第一个比它大的即i。环形相当于后面再贴了一遍相同元素，用取余来找出原来的位置。

Code

/*
 * @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;
    }
};

Analysis

时空复杂度都是O(N)