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
[LeetCode] Next Greater Element II 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 in the arr...
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 in the array, which means you could search ci...
503. Next Greater Element II Problem: 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 in the ...
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 in the array, which means you could search ci...
来自专栏 · LeetCode刷题记录 1 人赞同了该文章 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[...
For number 2 in the first array, there is no next greater number for it in the second array, so output -1. My solution: class Solution { public int[] nextGreaterElement(int[] nums1, int[] nums2) { int[] res = new int[nums1.length]; ...
作者是 在线疯狂 发布于 2017年2月5日 在LeetCode. 题目描述: LeetCode 496. Next Greater Element I 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...
publicintnextGreaterElement(int n){String value=String.valueOf(n);char[]digits=value.toCharArray();int i=digits.length-1;//找到小于右侧任意值的第一个正整数while(i>0){if(digits[i-1]<digits[i]){break;}i--;}if(i==0){return-1;}//找到该整数右侧大于该整数的最小整数int maxIndex=i,...
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; int[] result = new int[findLength]; ...