原题链接在这里:Majority Element I,Majority Element II 对于Majority Element I 来说,有多重解法. Method 1:最容易想到的就是用HashMap 计数,数值大于n/2(注意不是大于等于而是大于),就是返回值.Time O(n), Space O(n). Method 2: 用了sort,返回sort后array的中值即可.Time O(n*logn), Space O(1). Method 3: 维护个最常出现值,遇到相同count+
/* *JS时间戳比较大小:对于一组时间戳(开始时间~结束时间)和另一组时间戳进行比较,用于判断被比较时间戳组是否在要求范围内 *@param date1 date2(形如:'2015-01-01'类型字符串) */ function compareDate(date1,date2){ //对获得的时间戳区间与既定的时间戳进行比对 var baseDate1='2015-01-01'; var baseDate2='2015-03-31'; baseDate1=new Date(baseDate
In a array A of size 2N, there are N+1 unique elements, and exactly one of these elements is repeated N times. Return the element repeated N times. Example 1: Input: [1,2,3,3] Output: 3 Example 2: Input: [2,1,2,5,3,2] Output: 2 Example 3: Input: [5,1
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. 需要注意的是:这道题只是要找出多数元素,已经默认存在多数
如果直接使用Element做时间选择器,其规则(rules)不添加type:'date',会提示类型错误,处理这个需要规范值的类型为date. 时间格式化过滤器 import Vue from 'vue' import service from './service.js' import onlineIcon from '@/assets/img/map/tsp_green.png' import unlineIcon from '@/assets/img/map/tsp_gray.png' i
原题链接在这里:https://leetcode.com/problems/majority-element/ 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 ele
题目链接 Maximum Element 题意 现在有这一段求序列中最大值的程度片段: (假定序列是一个1-n的排列) int fast_max(int n, int a[]) { int ans = 0; int offset = 0; for (int i = 0; i < n; ++i) if (ans < a[i]) { ans = a[i]; offset = 0; } else { offset = offset + 1; if (offset == k) return ans
