classSolution:defmajorityElement(self, nums):""" :type nums: List[int] :rtype: int """majority_count =len(nums) //2fornum1innums: count =sum(1fornum2innumsifnum2 == num1)ifcount > majority_count:returnnum1 复杂度分析 时间复杂度:O(n2)O(n2),其中nn表示数组的长度,由于嵌套了两层...
AI代码解释 classSolution:defmajorityElement(self,nums):""":type nums:List[int]:rtype:int""" count=0major=0foriinnums:ifcount==0:major=iifmajor==i:count+=1else:count-=1returnmajor
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半次数)。 直观想法,HashMap,很慢 publicclassSolution {publicintmajorityElement(int[] nums) { Map map=newHashMap<Integer, Integer>();intlen ...
第一直觉是先排序把相同的元素都放到一起再说,因为主要元素的出现次数大于n/2,那么排序后最中间的元素一定是主要元素,不管怎么移动位置,最中间的都一定是它,所以可以很简单地完成代码啦。 代码(Java): 代码语言:javascript 代码运行次数:0 publicclassSolution{publicintmajorityElement(int[]nums){Arrays.sort(nums)...
public class Solution { public String reverseWords(String s) { // 使用trim()方法去除字符串前后的空格 s = s.trim(); // 使用StringBuilder来存储结果,避免频繁的字符串拼接操作 StringBuilder sb = new StringBuilder(); // 定义一个双指针,分别指向字符串的开头和末尾 int start = 0; int end = s...
About LeetCode Java solution Resources Readme Activity Stars 334 stars Watchers 24 watching Forks 126 forks Report repository Releases No releases published Packages No packages published Contributors 2 LjyYano Yano juhipariyal Languages Java 99.8% Shell 0.2% ...
我以前做过OJ，也支持了JAVA，JAVA普遍在内存和时间上都吃亏，因为JVM冷启动很慢，JVM本身也很占内存。
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Python,C++,Java都用了相同的哈希法 用hashmap记录每个元素出现的次数,最后遍历得到大于一半总数的那个数字 二.Python实现 classSolution:defmajorityElement(self,nums:List[int])->int:n=len(nums)dic={}foriinrange(n):dic[nums[i]]=dic.get(nums[i],0)+1foriinrange(n):ifdic[nums[i]]>n/2:retu...
LeetCode 刷题随手记 - 第一部分 前 256 题(非会员),仅算法题,的吐槽 https://leetcode.com/problemset/algorithms/...