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...
AI代码解释 classSolution{public:vector<int>nextGreaterElements(vector<int>&nums){int n=nums.size();stack<int>st;vector<int>res(n,-1);for(int i=2*n-2;i>=0;--i){while(!st.empty()&&nums[i%n]>=st.top()){st.pop();}res[i%n]=st.empty()?-1:st.top();st.push(nums[i%n]...
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. 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...
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 的下一个更大的元素是按数组遍历...
[LeetCode] 503. 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 ...
Leetcode 503 Next Greater Element II Leetcode 239 Sliding Window Maximum (唯一的单调队列题) 扫描线算法(Sweep Line) 基础知识:一个很巧妙的解决时间安排冲突的算法,本身比较容易些也很容易理解 常见题目: Leetcode 253 Meeting Room II(Meeting Room I也可以使用) Leetcode 1094 Car Pooling Leetcode 218 ...
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,...
0503 Next Greater Element II Go 56.5% Medium 0504 Base 7 46.2% Easy 0505 The Maze II 47.7% Medium 0506 Relative Ranks 50.5% Easy 0507 Perfect Number Go 35.5% Easy 0508 Most Frequent Subtree Sum Go 57.9% Medium 0509 Fibonacci Number Go 67.2% Easy 0510 Inorder Successor in BST...