packageleetcodeimport"math"funccommonChars(A[]string)[]string{cnt:=[26]int{}fori:=rangecnt{cnt[i]=math.MaxUint16}cntInWord:=[26]int{}for_,word:=rangeA{for_,char:=range[]byte(word){// compiler trick - here we will not allocate new memorycntInWord[char-'a']++}fori:=0;i<26...
思路:使用一个26个长度的int数组count,初始值设为整型最大值,循环处理A中的字符串,借助26个长度的int数组tem,将每个字符串中的字符出现次数记数,然后比较count和tem中对应位的元素值大小(字符出现次数),取较小(求交集)的一个重新赋值给count中的对应位,最后将count数组中大于0的数转成字符串添加到List中去。
classSolution{public:vector<string>commonChars(vector<string>&A){vector<string>r;intmin=0;for(inti=1;i<A.size();i++){if(A[i].length()<A[min].length()){min=i;}}set<char>st;for(inti=0;i<A[min].length();i++){st.insert(A[min][i]);}for(set<char>::iteratorit=st.begin(...
len(A)): tempcount = {} for i in A[word]: if i in tempcount: tempcount[i] += 1 else: tempcount[i] = 1 delelem = [] for i in dic.keys
今天介绍的是LeetCode算法题中Easy级别的第236题(顺位题号是1002)。给定仅由小写字母组成的字符串A,返回列表中所有字符串都有显示的字符的列表(包括重复字符)。例如,如果一个字符在所有字符串中出现3次但不是4次,则需要在最终答案中包含该字符三次。
Leetcode每日一题:1002.find-common-characters(查找常用字符串),思路:可用暴力法,找出最短字符串short_string,然后对里的每个字符c,都在A里面搜索字符c的数量,数量最小值就是c在所有字符串中出现的次数;也可用计数法,即求每个字符串之间字符数量的交集;暴力法
[LeetCode] 1002. Find Common Characters Given an arrayAof strings made only from lowercase letters, return a list of all characters that show up in all strings within the list (including duplicates). For example, if a character occurs 3 times in all strings but not 4 times, you need to...
Memory Usage: 38.1 MB, less than 100.00% of Java online submissions for Find Common Characters. class Solution { public List<String> commonChars(String[] A) { int n = A.length; int[][] cc = new int[n][26]; for (int i = 0; i < n; i++) { ...
方法: 统计每个字符串中26个英文字母出现的次数,然后取出现次数的最小值,该最小值即为每个字母重复的个数,实现原理参考下面的代码。 具体实现: classFindCommonCharacters{funcommonChars(A:Array<String>):List<String>{valletters=Array(A.size,{Array('z'-'Z',{0})})valresult=mutableListOf<String>()for...
Breadcrumbs leetcode /1002.find_common_characters / main.goTop File metadata and controls Code Blame 104 lines (92 loc) · 1.79 KB Raw package main import "fmt" // use hash table func commonChars(words []string) []string { resMap := make(map[rune]int) for _, word := range words...