LeetCode 496. Next Greater Element I

You are given two arrays (without duplicates) nums1 and nums2 where
nums1’s elements are 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, output -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. Note: All elements in nums1
and nums2 are unique. The length of both nums1 and nums2 would not
exceed 1000.

我的解法如下

class Solution {    public int[] nextGreaterElement(int[] nums1, int[] nums2) {        int i=0;        int j=0;        int []res=new int[nums1.length];        while(i<nums1.length){            if(nums1[i]==nums2[j]){                j=j+1;                while(j<nums2.length&&nums1[i]>=nums2[j]){                    j++;                }                if(j==nums2.length){                    res[i]=-1;                    j=0;                    i++;                }                else{                    res[i]=nums2[j];                    j=0;                     i++;                }            }            else{                j++;            }        }        return res;    }}

看discuss可以使用栈，先进后出，这样可以存储下一个最大的元素

  public int[] nextGreaterElement(int[] findNums, int[] nums) {        Map<Integer, Integer> map = new HashMap<>(); // map from x to next greater element of x        Stack<Integer> stack = new Stack<>();        for (int num : nums) {            while (!stack.isEmpty() && stack.peek() < num)                map.put(stack.pop(), num);            stack.push(num);        }           for (int i = 0; i < findNums.length; i++)            findNums[i] = map.getOrDefault(findNums[i], -1);        return findNums;    }