Leetcode-496的解题过程。

• 题目

  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.

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  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:

1. All elements in nums1 and nums2 are unique.
2. The length of both nums1 and nums2 would not exceed 1000.

• 解析

  给定两个没有重复元素的数组num1和num2,num1是num2的子集，对于num1中的每个元素a，要求找到在num2中a右边第一个比a大的元素，找不到返回-1。Easy题一道，没什么技术含量，根据题意直接搜索就行了。

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  public class Solution { public int[] NextGreaterElement(int[] findNums, int[] nums) { int[] result = new int[findNums.Length]; for (int i=0;i<findNums.Length;i++){ int index = Array.IndexOf(nums,findNums[i]); bool flag = false; for (int j=index+1;j<nums.Length;j++){ if (nums[j]>findNums[i]){ result[i] = nums[j]; flag = true; break; } } if (!flag){ result[i] = -1; } } return result; } }