Given two strings s and t, return true if t is an anagram of s, and false otherwise. An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the origi
LN : leetcode 242 Valid Anagram lc 242 Valid Anagram 242 Valid Anagram Given two strings s and t, write a function to determine if t is an anagram of s. For example, s = "anagram", t = "nagaram", return true. s = "rat", t = "car", return false. analysation## 利用类似于带...
关于题目中anagram的意思,结合给出的两个示例,大意就是两字符串使用的小写字母一样,但是每个字母所处的位置不是全都一样。此解法是将两字符串s、t转换为字符数组,然后将数组排序,最后比较两数组的元素是否相等,这里借助了工具类Arrays。 此解法的时间复杂度是O(nlog(n)),空间复杂度是O(n)。 publicbooleanisAn...
leetcode 242. Valid Anagram Given two stringssandt, write a function to determine iftis an anagram ofs. For example, s= "anagram",t= "nagaram", return true. s= "rat",t= "car", return false. Note: You may assume the string contains only lowercase alphabets. Follow up: What if the...
LeetCode: 242. Valid Anagram LeetCode: 242. Valid Anagram 题目描述 Given two strings s and t , write a function to determine if t is an anagram of s. Example 1: Input: s = "anagram", t = "nagaram"...
242. 有效的字母异位词 Valid Anagram LeetCodeCN 第242题链接 第一种方法:对两个字符串排序后对比 classSolution:defisAnagram(self,s:str,t:str)->bool:returnsorted(s)==sorted(t) 第二种方法:用哈希表对字符串内每个字符计数,最后比对哈希表,这里用dict实现 ...
LeetCode.jpg 有效的字母异位词 给定两个字符串 s 和 t ,编写一个函数来判断 t 是否是 s 的一个字母异位词。 示例1: 输入: s = "anagram", t = "nagaram" 输出: true 示例2: 输入: s = "rat", t = "car" 输出: false 说明: 你可以假设字符串只包含小写字母。
Can you solve this real interview question? Valid Anagram - Given two strings s and t, return true if t is an anagram of s, and false otherwise. Example 1: Input: s = "anagram", t = "nagaram" Output: true Example 2: Input: s = "rat", t = "car" O
Valid Anagram(leetcode242) Given two strings s and t , write a function to determine if t is an anagram of s. Example 1: Input: s = “anagram”, t = “nagaram” Output: true Example 2: Input: s = “rat&...猜你喜欢LeetCode 242. Valid Anagram 242. Valid Anagram......
classSolution {public:boolisAnagram(strings,stringt) {if(s.size() != t.size())returnfalse;intm[26] = {0};for(inti =0; i < s.size(); ++i) ++m[s[i] -'a'];for(inti =0; i < t.size(); ++i) {if(--m[t[i] -'a'] <0)returnfalse; ...