496. Next Greater Element I

难度: Easy

题目描述

The next greater element of some element x in an array is the first greater element that is to the right of x in the same array.

You are given two distinct 0-indexed integer arrays nums1 and nums2, where nums1 is a subset of nums2.

For each 0 <= i < nums1.length, find the index j such that nums1[i] == nums2[j] and determine the next greater element of nums2[j] in nums2. If there is no next greater element, then the answer for this query is -1.

Return an array ans of length nums1.length such that ans[i] is the next greater element as described above.

 

Example 1:

Input: nums1 = [4,1,2], nums2 = [1,3,4,2]
Output: [-1,3,-1]
Explanation: The next greater element for each value of nums1 is as follows:
- 4 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1.
- 1 is underlined in nums2 = [1,3,4,2]. The next greater element is 3.
- 2 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1.

Example 2:

Input: nums1 = [2,4], nums2 = [1,2,3,4]
Output: [3,-1]
Explanation: The next greater element for each value of nums1 is as follows:
- 2 is underlined in nums2 = [1,2,3,4]. The next greater element is 3.
- 4 is underlined in nums2 = [1,2,3,4]. There is no next greater element, so the answer is -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?

解题思路

代码实现

解决方案

java

class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {

        Map<Integer, Integer> elementIndexMap = new HashMap<>();
        for (int i = 0; i < nums2.length; i++) {
            elementIndexMap.put(nums2[i], i);
        }
        int[] resArray = new int[nums1.length];

        for (int i = 0; i < nums1.length; i++) {
            int targetIndex = elementIndexMap.get(nums1[i]);
            int tempRes = 0;
            for (int j = targetIndex + 1; j < nums2.length; j++) {
                if (nums2[j] > nums1[i]) {
                    tempRes=nums2[j];
                    break;
                }
            }
            if (tempRes == 0) {
                tempRes = -1;
            }
            resArray[i] = tempRes;
        }

        return resArray;

    }
}