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);6
Leetcode 556. Next Greater Element III 2. Solution **解析:**Version 1,先将数字n变为字符数组,要找最小的大于n的数,则应该从右往左开始,依次寻找第i位字符右边的大于当前字符的最小数字,然后互换二者位置,由于新数字的第i位字符大于n中的第i位字符,因此新数字i位之后的字符应该从小到大排列,这样可以保...
Next Greater Element I? Leetcode 496. Next Greater Element I的时间复杂度是多少? 1. Description 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 代码语言:...
class Solution { public int[] nextGreaterElement(int[] nums1, int[] nums2) { int l1 = nums1.length, l2 = nums2.length; int[] res = new int[l1]; for(int i = 0; i < l1; i++){ for(int j = 0; j < l2; j++){ if(nums2[j] != nums1[i]) //locate the number in...
Can you solve this real interview question? Next Greater Element II - Given a circular integer array nums (i.e., the next element of nums[nums.length - 1] is nums[0]), return the next greater number for every element in nums. The next greater number of
503. Next Greater Element II 难度:m class Solution: def nextGreaterElements(self, nums: List[int]) -> List[int]: if not nums: return [] stack = [] res = [-1]*len(nums) for i in range(len(nums)): while stack and nums[stack[-1]]<nums[i]: res[stack.pop()] = nums[i]...
【LeetCode】496. Next Greater Element I 解题报告(Python & C++),【LeetCode】496.NextGreaterElementI解题报告标签(空格分隔):LeetCode题目地址:https://leetcode.com/problems/self-dividing-numbers/description/题目描述:Youaregiventwoarrays(withoutduplica
题目描述: LeetCode 556. Next Greater Element III Given a positive 32-bit integer n, you need to find the smallest 32-bit integer which has exactly the same digits existing in the integer n and is greater in value than n. If no such positive 32-bit integ
[leetcode] 503. Next Greater Element II Description 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 greater number to its traversing-order...
Next Greater Element I ...[LeetCode] 496. Next Greater Element I 题目内容 https://leetcode-cn.com/problems/next-greater-element-i/ 给定两个没有重复元素的数组 nums1 和 nums2 ,其中nums1 是 nums2 的子集。找到 nums1 中每个元素在 nums2 中的下一个比其大的值。 ......