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(…

169. 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. 找出数列中个…

169. 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. 思路1:ha…

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. 题目标签:Array 忘记说了,特地回来补充,今天看完<…

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. 思路: Find k different element…

题目描述 给定一个大小为 n 的数组,找到其中的众数.众数是指在数组中出现次数大于 ⌊ n/2 ⌋ 的元素. 你可以假设数组是非空的,并且给定的数组总是存在众数. 示例 1: 输入: [3,2,3] 输出: 3 示例 2: 输入: [2,2,1,1,1,2,2] 输出: 2 思路 思路一: 利用哈希表的映射,储存数组中的数字以及它们出现的次数,当众数出现时,返回这个数字. 思路二: 因为众数是出现次数大于n/2的数字,所以排序之后中间的那个数字一定是众数.即nums[n/2]为众数.但是在计算比…

1. 题目描述Description Link: https://leetcode.com/problems/majority-element/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-e…

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. Example 1: Input: [3,2,3] Ou…

169. 多数元素 给定一个大小为 n 的数组,找到其中的多数元素.多数元素是指在数组中出现次数大于 ⌊ n/2 ⌋ 的元素. 你可以假设数组是非空的,并且给定的数组总是存在多数元素. 示例 1: 输入: [3,2,3] 输出: 3 示例 2: 输入: [2,2,1,1,1,2,2] 输出: 2 class Solution { public int majorityElement(int[] nums) { int count = 1; int maj = nums[0]; for (int…

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. 给定一个数组,求其中权制最大的元素,(该元素出现超过了一…

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. 分析: 遍历数组,每当发现一对儿不相同的element时…

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. 这一题可以用排序之后查看序列正中间那个元素的方法来解.但…

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. 解题思路: 编程之美P130(寻找发帖水王)原题,如果删…

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…

题目描述: 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. 解题思路: 每找出两个不同的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. 思路: 找到一个数组中出现次数超过一半的数.排序.哈希等…

题目要求 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. 题目分析及思路 给定一个长度为n的数组,找到m…

问题描述 给定一个大小为 n 的数组,找到其中的众数.众数是指在数组中出现次数大于 ⌊ n/2 ⌋ 的元素. 你可以假设数组是非空的,并且给定的数组总是存在众数. 示例 1: 输入: [3,2,3] 输出: 3 示例 2: 输入: [2,2,1,1,1,2,2] 输出: 2 解决方案 class Solution: # two pass + dictionary def majorityElement1(self, nums): dic = {} for num in nums: dic[num…

题目: 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. 也就是找数组中出现次数大于一半的数字,题目保证这…

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. Hide Tags: Divide and Conquer…

题目: 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. 题解:运用多数投票算法的思路来解:从头到尾遍历数…

在一个长度为n的数组中找出出现次数超过(n+1)/2次的数 说明请参考编程之美中的2.3 class Solution { public: int majorityElement(vector<int>& nums) { int candidate; int ntimes,i; ; i < nums.size(); ++i){ ){ candidate = nums[i],ntimes = ; } else{ if(candidate == nums[i]) ntimes ++;…

给定一个大小为 n 的数组,找到其中的多数元素.多数元素是指在数组中出现次数大于 ⌊ n/2 ⌋ 的元素. 你可以假设数组是非空的,并且给定的数组总是存在多数元素. 示例 1: 输入: [3,2,3] 输出: 3 示例 2: 输入: [2,2,1,1,1,2,2] 输出: 2 Code:sort.hash.BM投票.随机数.位运算 class Solution { public: // 先排序 直接返回 N/2位置的元素 (无论N是奇数还是偶数) int majorityElement(vect…

public static int majorityElement(int[] nums) { int num = nums[0], count = 1; for(int i=1;i<nums.length;i++){ if(nums[i] == num) { count++; } else if(--count < 0) { num = nums[i]; count = 1; } } return num; }…

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. 思路: [LeetCode 169]Majority Element 的拓展,这回要求的是出现次数超过三分之一次的数字咯,动动我们的大脑思考下,这样的数最多会存在几个呢,当然是2个嘛.因此,接着上一题的方…

刷题备忘录,for bug-free leetcode 396. Rotate Function 题意: Given an array of integers A and let n to be its length. Assume Bk to be an array obtained by rotating the array A k positions clock-wise, we define a "rotation function" F on A as follow: F(k…

刷题备忘录,for bug-free 招行面试题--求无序数组最长连续序列的长度,这里连续指的是值连续--间隔为1,并不是数值的位置连续 问题: 给出一个未排序的整数数组,找出最长的连续元素序列的长度. 如: 给出[100, 4, 200, 1, 3, 2], 最长的连续元素序列是[1, 2, 3, 4].返回它的长度:4. 你的算法必须有O(n)的时间复杂度 . 解法: 初始思路 要找连续的元素,第一反应一般是先把数组排序.但悲剧的是题目中明确要求了O(n)的时间复杂度,要做一次排序,是不能达…

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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] Output: [1,2] 169. Maj…

leetcode探索中级答案汇总: https://leetcode-cn.com/explore/interview/card/top-interview-questions-medium/ 1)数组和字符串: leetcode 15 三数之和(medium)排序+双指针 leetcode73 矩阵置零 (medium) 空间节省技巧 leetcode 49 字母异位词分组(medium)排序+哈希 leetcode 3 无重复字符的最长子串(medium) DP leetcode5 最长回文…