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
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...
https://leetcode.cn/problems/next-greater-element-ii 给定一个循环数组 nums ( nums[nums.length - 1] 的下一个元素是 nums[0] ),返回 nums 中每个元素的 下一个更大元素 。 数字x 的 下一个更大的元素 是按数组遍历顺序,这个数字之后的第一个比它更大的数,这意味着你应该循环地搜索它的下一个更...
LeetCode——503. 下一个更大元素 II[Next Greater Element II][中等]——分析及代码[Java] 一、题目 二、分析及代码 1. 单调栈 (1)思路 (2)代码 (3)结果 三、其他 一、题目 给定一个循环数组(最后一个元素的下一个元素是数组的第一个元素),输出每个元素的下一个更大元素。数字 x 的下一个更大...
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] 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...
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...
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...
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,...
[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。 ......