LeetCode 496. Next Greater Element I

最新推荐文章于 2025-12-16 18:10:48 发布

转载最新推荐文章于 2025-12-16 18:10:48 发布 · 210 阅读

文章标签：

#leetcode

LeetCode 同时被 2 个专栏收录

25 篇文章

订阅专栏

堆栈、HashMap

6 篇文章

订阅专栏

本文介绍了一种算法问题的解决方案：给定两个数组，找到第一个数组中每个元素在第二个数组中对应的下一个更大元素。通过两种方法实现，一种是简单遍历比较，另一种利用栈的数据结构优化查找过程。

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;
    }