Can you solve this real interview question? Next Greater Element I - The next greater element of some element x in an array is the first greater element that is to the right of x in the same array. You are given two distinct 0-indexed integer arrays num
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...
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:...
For example [5, 4, 3, 2, 1, 6] then the greater number 6 is the next greater element for all previous numbers in the sequence We use a stack to keep a decreasing sub-sequence, whenever we see a number x greater than stack.peek() we pop all elements less than x and for all th...
[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。 ......
class Solution1 { public: vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) { int findNumsSize=findNums.size(); int numsSize=nums.size(); vector<int> ret(findNumsSize); for(int i=0; i<findNumsSize; ++i){ ...
Output: [-1,3,-1] 2、代码实现 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; ...
简介:LeetCode之Next Greater Element I 1、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...
## 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...
stk便成了从顶向下增大的栈. 最后利用hm把findNums的每个元素对应的next greater element找出来. Time Complexity: O(n), n = nums2.length. Space: O(n). AC Java: 1classSolution {2publicint[] nextGreaterElement(int[] nums1,int[] nums2) {3HashMap<Integer, Integer> hm =newHashMap<Integer,...