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1、題目

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.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.

給你兩個(gè)數(shù)組，為數(shù)組1和數(shù)組2，數(shù)組1為數(shù)組2的子集。找出數(shù)組1的每一個(gè)元素在數(shù)組2中對(duì)應(yīng)的元素a,然后找到元素a后側(cè)第一個(gè)比a大的數(shù)構(gòu)成一個(gè)數(shù)組，即是我們需要的答案。如果不存在，則為-1。

Input: nums1 = [4,1,2], nums2 = [1,3,4,2]. Output: [-1,3,-1]
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2、代碼實(shí)現(xiàn)

public class Solution {public int[] nextGreaterElement(int[] findNums, int[] nums) {if (findNums == null || nums == null) {return null;}int findLength = findNums.length;int numsLength = nums.length;int[] result = new int[findLength];boolean flag = false;for (int i = 0; i < findLength; ++i) {for (int j = 0; j < numsLength; ++j) {if (findNums[i] == nums[j]) {if (j + 1 == numsLength) {result[i] = -1;} else {for (int k = j + 1; k < numsLength; ++k) {if (nums[j] < nums[k]) { result[i] = nums[k];flag = true;break;}}if (!flag) {result[i] = -1;}}flag = false;}}}return result; } }
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3、總結(jié)我遇到的問(wèn)題

我一開(kāi)始寫(xiě)這里的代碼的時(shí)候 for (int k = j + 1; k < numsLength; ++k) {if (nums[j] < nums[k]) { result[i] = nums[k];flag = true;break;}} 忘記了寫(xiě)break; 然后導(dǎo)致第一個(gè)元素符合條件了，但是后面也有符合條件的，導(dǎo)致result[i]，取到的元素是最后面符合條件的元素，就錯(cuò)了，所以，這個(gè)時(shí)候條件是說(shuō)，得到第一個(gè)大于前面的數(shù) 就可以了，我們需要用break跳出循環(huán)，防穿透，防止繼續(xù)便利后面的元素，所以以后編程，看到符合后面第一個(gè)元素滿足要求的時(shí)候,要記得用break;形成條件反射。

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總結(jié)

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