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 ar…

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 ar…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 暴力解法 单调递减栈 日期 题目地址:https://leetcode.com/problems/minesweeper/description/ 题目描述 Given a circular array (the next element of the last element is the first element of the array),…

▶ 给定一个数组与它的一个子列,对于数组中的一个元素,定义它右边第一个比他大的元素称为他的后继,求所给子列的后继构成的数组 ▶ 第 496 题,规定数组最后一个元素即数组最大元素的后继均为 -1 ● 自己的版本,12 ms,最快的解法算法与之相同 class Solution { public: vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) { const…

https://leetcode.com/problems/next-greater-element-ii/description/ class Solution { public: vector<int> nextGreaterElements(vector<int>& nums) { stack<int> st; int n = nums.size(); * n - ; i >= n; i--) { int cur = nums[i % n]; whi…

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大的数.如果不存在,则输出 -1.示例 1:输入: [1,2,1]输出: [2,-1,2]解释: 第一个 1 的下一个更大的数是 2:数字 2 找不到下一个更大的数: 第二个 1 的下一个最大的数需要循环搜索,结果也是 2.注意: 输入数组的长度不会超过 10000.详见:https://…

503. 下一个更大元素 II 503. Next Greater Element II 题目描述 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大的数.如果不存在,则输出 -1. LeetCode503. Next Greater Element II中等 示例 1: 输入: [1,2,1] 输出: [2,-1,2] 解释: 第一个…

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 ar…

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 ar…

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 nums1 is t…

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 nand is greater in value than n. If no such positive 32-bit integer exists, you need to return -1. Example 1: In…

556. 下一个更大元素 III 556. Next Greater Element III 题目描述 给定一个 32 位正整数 n,你需要找到最小的 32 位整数,其与 n 中存在的位数完全相同,并且其值大于 n.如果不存在这样的 32 位整数,则返回-1. LeetCode556. Next Greater Element III中等 示例 1: 输入: 12 输出: 21 示例 2: 输入: 21 输出: -1 示例 3: 输入: 12443322 输出: 13222344 Java 实现…

496. 下一个更大元素 I 496. Next Greater Element I 题目描述 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. nums1 中数字 x 的下一个更大元素是指 x 在 nums2 中对应位置的右边的第一个比 x 大的元素.如果不存在,对应位置输出 -1. 每日一算法2019/6/7Day 35LeetCode496. Next Greater Ele…

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 integer exists, you need to return -1. Example 1: 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 of nums2. The Next Greater Number of a number x in nums1 is t…

2018-09-24 21:52:38 一.Next Greater Element I 问题描述: 问题求解: 本题只需要将nums2中元素的下一个更大的数通过map保存下来,然后再遍历一遍nums1即可. public int[] nextGreaterElement(int[] nums1, int[] nums2) { int[] res = new int[nums1.length]; Map<Integer, Integer> map = new HashMap<>()…

原题链接在这里:https://leetcode.com/problems/next-greater-element-iii/description/ 题目: 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 nand is greater in value than n.…

［抄题］: 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 nums…

一个数组里有一个数重复了n/2多次,找到 思路:既然这个数重复了一半以上的长度,那么排序后,必然占据了 a[n/2]这个位置. class Solution { public: int majorityElement(vector<int>& nums) { sort(nums.begin(),nums.end()); return nums[nums.size()/2]; } }; 线性解法:投票算法,多的票抵消了其余人的票,那么我的票一定还有剩的. int majority; in…

一:Majority Element Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times. You may assume that the array is non-empty and the majority element always exist in the array. class Sol…

169. Majority Element 求超过数组个数一半的数 可以使用hash解决,时间复杂度为O(n),但空间复杂度也为O(n) class Solution { public: int majorityElement(vector<int>& nums) { unordered_map<int,int> count; int n=nums.size(); ;i<n;i++){ ) return nums[i]; } ; } }; 使用投票法,时间复杂度为O(…

寻找多数元素这一问题主要运用了:Majority Vote Alogrithm(最大投票算法)1.Majority Element 1)description Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times. You may assume that the array is non-empty…

Majority Element II Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. The algorithm should run in linear time and in O(1) space. Hint: How many majority elements could it possibly have? Do you have a better hint…

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 nums1 is t…

LeetCode--Next Greater Element I Question 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…

https://leetcode.com/problems/next-greater-element-i/#/description 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 p…

LeetCode169. Majority Element Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times. (Easy) You may assume that the array is non-empty and the majority element always exist in th…

[LeetCode]556. Next Greater Element III 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址:https://leetcode.com/problems/next-greater-element-iii/description/ 题目描述: Given a positive 32-bit integer n, you need…

Majority Element II Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. Note: The algorithm should run in linear time and in O(1) space. Example 1: Input: [3,2,3] Output: [3] Example 2: Input: [1,1,1,3,3,2,2,2] Ou…

Question 496. Next Greater Element I Solution 题目大意:给你一个组数A里面每个元素都不相同.再给你一个数组B,元素是A的子集,问对于B中的每个元素,在A数组中相同元素之后第一个比它的元素是多少. 思路:把nums1中的元素存储到一个map里,遍历nums2,如果能从map中取到值,就遍历nums2中后续元素的值并与当前元素做比较,如果存在比当前元素大的值就取该值(第一个),否则返回-1. Java实现: public int[] nextGreate…