README_EN.md

496. Next Greater Element I

中文文档

Description

You are given two integer arrays nums1 and nums2 both of unique elements, where nums1 is a subset of nums2.

Find all the next greater numbers for nums1's elements in the corresponding places of nums2.

The Next Greater Number of a number x in nums1 is the first greater number to its right in nums2. If it does not exist, return -1 for this number.

 

Example 1:

Input: nums1 = [4,1,2], nums2 = [1,3,4,2]
Output: [-1,3,-1]
Explanation:
For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1.
For number 1 in the first array, the next greater number for it in the second array is 3.
For number 2 in the first array, there is no next greater number for it in the second array, so output -1.

Example 2:

Input: nums1 = [2,4], nums2 = [1,2,3,4]
Output: [3,-1]
Explanation:
For number 2 in the first array, the next greater number for it in the second array is 3.
For number 4 in the first array, there is no next greater number for it in the second array, so output -1.

 

Constraints:

• 1 <= nums1.length <= nums2.length <= 1000
• 0 <= nums1[i], nums2[i] <= 104
• All integers in nums1 and nums2 are unique.
• All the integers of nums1 also appear in nums2.

 

Follow up: Could you find an O(nums1.length + nums2.length) solution?

Solutions

Python3

class Solution:
    def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
        mapper = dict()
        stack = []
        for num in nums2:
            while stack and stack[-1] < num:
                mapper[stack.pop()] = num
            stack.append(num)
        res = []
        for num in nums1:
            res.append(mapper.get(num, -1))
        return res

Java

class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        Deque<Integer> stack = new ArrayDeque<>();
        Map<Integer, Integer> map = new HashMap<>();
        for (int num : nums2) {
            while (!stack.isEmpty() && stack.peek() < num) {
                map.put(stack.pop(), num);
            }
            stack.push(num);
        }
        int n = nums1.length;
        int[] res = new int[n];
        for (int i = 0; i < n; ++i) {
            res[i] = map.getOrDefault(nums1[i], -1);
        }
        return res;
    }
}

...