对于t, 同样将其遍历,对每个出现的字符计数减一。 如果s和t是anagram , 那么最后的charcount数组中所有字符的计数都应该是0, 否则就不是anagram。 [Code] 1: class Solution { 2: public: 3: boolisAnagram(string s, string t) { 4: vector<int>charcount(26,0); 5:for(int i =0;i< s.length(...
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. 解法一:排序后判相等 classSolution {publ...
class Solution { public boolean isAnagram(String s, String t) { if (s.length() != t.length()) { return false; } char[] str1 = s.toCharArray(); char[] str2 = t.toCharArray(); Arrays.sort(str1); Arrays.sort(str2); return Arrays.equals(str1, str2); } } 复杂度分析 时间复...
class Solution(object): def isAnagram(self, s, t): """ :type s: str :type t: str :rtype: bool """ if len(s) != len(t): return False hash_map = {} for i in range(len(s)): hash_map[s[i]] = hash_map.get(s[i], 0) + 1 hash_map[t[i]] = hash_map.get(t[...
242. Valid Anagram 首先,要先弄清楚什么是anagram,anagram是指由颠倒字母顺序组成的单词。 解决方法一种是排序,一种是哈希表。 solution: class Solution { public: bool isAnagram(string s, string t) { if(s.length()!=t.length()) return false; ...
public class Solution { public boolean isAnagram(String s, String t) { char[] sArr = s.toCharArray(); char[] tArr = t.toCharArray(); Arrays.sort(sArr); Arrays.sort(tArr); return String.valueOf(sArr).equals(String.valueOf(tArr)); ...
242 Valid Anagram Easy Go 246 Strobogrammatic Number 🔒 Easy Go 253 Meeting Rooms II 🔒 Medium Go 278 First Bad Version Easy 283 Move Zeroes Easy Go 328 Odd Even Linked List Medium Go 334 Increasing Triplet Subsequence Medium Go 338 Counting Bits Easy Go 340 Longest Substring with At ...
0242 Valid Anagram Go 62.7% Easy 0243 Shortest Word Distance 64.9% Easy 0244 Shortest Word Distance II 60.7% Medium 0245 Shortest Word Distance III 57.5% Medium 0246 Strobogrammatic Number 47.6% Easy 0247 Strobogrammatic Number II 51.4% Medium 0248 Strobogrammatic Number III 41.7% Hard...
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 ...
20 Valid Parentheses 232 Implement Queue using Stacks【C】【My C solution】 225 Implement Stack using Queues【C】 Heap 703 Kth Largest Element in a Stream 239 Sliding Window Maximum Hash table 242 Valid Anagram 1 Two Sum【C】 15 3Sum 18 4Sum Tree, Binary tree, Binary search tree 98 Valid...