README_EN.md

503. Next Greater Element II

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Description

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

Solutions

Python3

class Solution:
    def nextGreaterElements(self, nums: List[int]) -> List[int]:
        n = len(nums)
        res = [-1] * n
        stk = []
        for i in range(n << 1):
            while stk and nums[stk[-1]] < nums[i % n]:
                res[stk.pop()] = nums[i % n]
            stk.append(i % n)
        return res

Java

class Solution {
    public int[] nextGreaterElements(int[] nums) {
        int n = nums.length;
        int[] res = new int[n];
        Arrays.fill(res, -1);
        Deque<Integer> stk = new ArrayDeque<>();
        for (int i = 0; i < (n << 1); ++i) {
            while (!stk.isEmpty() && nums[stk.peek()] < nums[i % n]) {
                res[stk.pop()] = nums[i % n];
            }
            stk.push(i % n);
        }
        return res;
    }
}

JavaScript

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var nextGreaterElements = function (nums) {
    let n = nums.length;
    let stack = [];
    let res = new Array(n).fill(-1);
    for (let i = 0; i < 2 * n; i++) {
        let cur = nums[i % n];
        while (stack.length > 0 && nums[stack[stack.length - 1]] < cur) {
            res[stack.pop()] = cur;
        }
        stack.push(i % n);
    }
    return res;
};

C++

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

Go

func nextGreaterElements(nums []int) []int {
	n := len(nums)
	res := make([]int, n)
	for i := range res {
		res[i] = -1
	}
	var stk []int
	for i := 0; i < (n << 1); i++ {
		for len(stk) > 0 && nums[stk[len(stk)-1]] < nums[i%n] {
			res[stk[len(stk)-1]] = nums[i%n]
			stk = stk[:len(stk)-1]
		}
		stk = append(stk, i%n)
	}
	return res
}

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