数组和矩阵（3）——Next Greater Element I

最新推荐文章于 2025-05-06 13:57:55 发布

转载最新推荐文章于 2025-05-06 13:57:55 发布 · 227 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/-1307/p/6930606.html

本文介绍了一种算法问题——寻找给定数组中每个元素的下一个更大的元素，并提供了两种解决方案：一种是简单的暴力法，另一种则是利用栈和哈希映射来提高效率。通过这两种方法，可以有效地解决LeetCode上的“下一个更大元素I”问题。

https://leetcode.com/problems/next-greater-element-i/#/description

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.

1.暴力

 1 public class Solution {
 2     public int[] nextGreaterElement(int[] findNums, int[] nums) {
 3         int len = findNums.length;
 4         int[] res = new int[len];
 5         if(len > nums.length) {
 6             return res;
 7         }
 8         for(int i=0; i<len; i++) {
 9             boolean flag = false;
10             boolean k = false;
11             for(int j=0; j<nums.length; j++) {
12                 if(findNums[i] == nums[j]) {
13                     flag = true;
14                     continue;
15                 }
16                 if(flag == true && nums[j] > findNums[i]) {
17                     res[i] = nums[j];
18                     k = true;
19                     break;
20                 }
21             }
22             if(k == false) {
23                 res[i] = -1;
24             }
25         }
26         return res;
27     }
28 }

2.集合

 1     public int[] nextGreaterElement(int[] findNums, int[] nums) {
 2         Map<Integer, Integer> map = new HashMap<>(); // map from x to next greater element of x
 3         Stack<Integer> stack = new Stack<>();
 4         for (int num : nums) {
 5             while (!stack.isEmpty() && stack.peek() < num)
 6                 map.put(stack.pop(), num);
 7             stack.push(num);
 8         }   
 9         for (int i = 0; i < findNums.length; i++)
10             findNums[i] = map.getOrDefault(findNums[i], -1);
11         return findNums;
12     }