1classSolution {2publicint[] nextGreaterElement(int[] nums1,int[] nums2) {3int[] res =newint[nums1.length];4Map<Integer,Integer> map =newHashMap<>();5for(inti = 0 ; i < nums2.length ; i ++) map.put(nums2[i],i);6for(inti = 0 ; i < nums1.length ; i ++){7for(in...
想到了用哈希表存这个数的位置,但是没有想到可以直接用哈希表存next great,用栈存还没找到的数,没遍历一个数就考察栈中的元素小,小的话,这个数就是栈中数的next great,栈中的数肯定是下大上小。 publicint[] nextGreaterElement(int[] nums1,int[] nums2) {/*通过map建立当前元素和其next great的映射 ...
题目地址:https://leetcode.com/problems/minesweeper/description/ 题目描述 Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element. The Next Greater Number of a number x is the first grea...
LeetCode题目:503. Next Greater Element II Given acircular array (the next element of the last element is the first element of the array),print the Next Greater Number for every element. The Next Greater Number of a number x is the first greater number to its traversing-order next in the...
Leetcode 503. Next Greater Element II 1. Description 2. Solution **解析:**Version 1,由于元素不是唯一的,需要循环查找,因此先将nums复制一遍,通过循环每次都查找当前元素之后的n-1位数字。Version 2通过使用栈来寻找满足条件的结果,栈中保持是数字的索引位置,由于需要循环查找,因此需要查找两次nums,并且第二...
Next Greater Element I 2. Solution 解析:Version 1,由于元素是唯一的,通过循环找出每个nums2中的满足条件结果保存到字典中,遍历nums1,获得结果。Version 2通过使用栈来寻找满足条件的结果,减少搜索时间。 Version 1 classSolution:defnextGreaterElement(self,nums1:List[int],nums2:List[int])->List[int]:stat...
## LeetCode 496fromtypingimportListclassSolution:defnextGreaterElement(self,nums1:List[int],nums2:List[int])->List[int]:res={}stack=[]foriinnums2[::-1]:## 从 num2 最右边的元素开始遍历whilestackandi>=stack[-1]:## 如果栈非空,且该元素比栈顶(列表最右边)元素大stack.pop()## 清空这个...
/* * @lc app=leetcode id=503 lang=cpp * * [503] Next Greater Element II * * https://leetcode.com/problems/next-greater-element-ii/description/ * * algorithms * Medium (51.85%) * Likes: 791 * Dislikes: 48 * Total Accepted: 57.8K * Total Submissions: 111.2K * Testcase Example:...
下面将使用两种方法来解题,一种是暴力破解法,一种是Next Greater Number问题解法。 2.1、暴力破解法 从头开始遍历nums1,每次遍历从nums2中找到对应元素,然后从该元素的下一个元素开始依次比较,找出大于该值的第一个元素即可。 【代码实现】 class Solution {public int[] nextGreaterElement(int[] nums1, int[]...
Given a circular integer arraynums(i.e., the next element ofnums[nums.length - 1]isnums[0]), returnthenext greater numberfor every element innums. Thenext greater numberof a numberxis the first greater number to its traversing-order next in the array, which means you could search circul...