Leetcode No.169 Majority Element(c++实现) 1. 题目 1.1 英文题目 Given an array nums of size n, return the majority element. The majority element is the element that appears more than ⌊n / 2⌋ times. You may assume that the majority element always exists in the array. 1.2 中文题目 ...
classSolution{publicintmajorityElement(int[] nums){intmajorityCount=nums.length /2;for(intnum1 : nums) {intcount=0;for(intnum2 : nums) {if(num2 == num1) { ++count; } }if(count > majorityCount) {returnnum1; } }thrownewIllegalArgumentException("The array does not contain a majority...
169 classSolution:defmajorityElement(self,nums:List[int])->int:current=0count=0foriinrange(len(nums)):ifnums[i]==nums[current]:count+=1else:count-=1ifcount==0:current=icount=1returnnums[current] 229 class Solution: def majorityElement(self, nums: List[int]) -> List[int]: current1,...
You may assume that the array is non-empty and the majority element always exist in the array. 就是使用HashMap直接计数。 代码如下: import java.util.HashMap; import java.util.Map; public class Solution { public int majorityElement(int[] nums) { Map<Integer, Integer> map=new HashMap<Integ...
活动作品【300题刷题挑战】leetcode力扣剑指 Offer 39. 数组中出现次数超过一半的数字 majorityElement 第二百四十九题 | 数学 30播放 ·1弹幕2021-08-11 23:27:01 主人,未安装Flash插件,暂时无法观看视频,您可以… 未经作者授权,禁止转载 力扣300题刷题挑战 第二百四十九题 数学 需要源码的请看: https://git...
题目地址:https://leetcode-cn.com/problems/check-if-a-number-is-majority-element-in-a-sorted-array/ 题目描述 Given an array nums sorted in non-decreasing order, and a number target, return True if and only if target is a majority element. ...
这道题出现在了王道的《2019数据结构考研复习指导》的18页,LeetCode中也有这道题。题目大意是:给定一个长度为n的数组,我们定义"主元素"为数组中出现次数超过⌊n/2⌋的元素。 Description Given an array of size n, find the majority element. The majority element is the element that appears more than...
169.Majority Element 这是leetCode第169题 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. ...
class Solution: def majorityElement(self, nums): """ :type nums: List[int] :rtype: List[int] """ if not nums: return [] # 候选参数1,候选参数2,初始化可以为任意值,只要保证num1 != num2 即可 num1, num2 = 0, 1 count1, count2 = 0, 0 ...
30 changes: 30 additions & 0 deletions 30 leetcode-x/find-majority-element/src/main/java/Solution.java Original file line numberDiff line numberDiff line change @@ -32,4 +32,34 @@ public static int majorityElement(int[] nums) { } return -1; } /** * 摩尔投票算法 * @param nums ...