496. Next Greater Element I
Easy
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
andnums2
are unique. - The length of both
nums1
andnums2
would not exceed 1000.
package leetcode.easy; public class NextGreaterElementI { private static void print_arr(int[] nums) { for (int num : nums) { System.out.print(num + " "); } System.out.println(); } public int[] nextGreaterElement(int[] nums1, int[] nums2) { int[] arr = new int[nums1.length]; for (int i = 0; i < nums1.length; i++) { int current = nums1[i]; int j = 0; while (nums2[j] != current) { j++; } for (j = j + 1; j < nums2.length; j++) { if (nums2[j] > current) { arr[i] = nums2[j]; break; } } if (j == nums2.length) { arr[i] = -1; } } return arr; } @org.junit.Test public void test1() { int[] nums1 = { 4, 1, 2 }; int[] nums2 = { 1, 3, 4, 2 }; print_arr(nextGreaterElement(nums1, nums2)); } @org.junit.Test public void test2() { int[] nums1 = { 2, 4 }; int[] nums2 = { 1, 2, 3, 4 }; print_arr(nextGreaterElement(nums1, nums2)); } }
原文地址:https://www.cnblogs.com/denggelin/p/12117373.html
时间: 2024-11-06 07:48:01