LeetCode-Valid Anagram
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数两个string的各个字母个数,然后比较两个计数的数组每一位是否一样
public class Solution { public boolean isAnagram(String s, String t) { int [] sArr = new int [26]; int [] tArr = new int [26]; if (s.length() != t.length() ) return false; for ( int i = 0; i < s.length(); i ++){ char c = s.charAt(i); int pos = c - 'a'; sArr[ pos ] ++; char c2 = t.charAt(i); int pos2 = c2 - 'a'; tArr[ pos2 ] ++; } for (int i = 0; i < 26; i ++){ if ( sArr[i] != tArr[i]) return false; } return true; }}
简便做法是 只用一个array 第一个s里面的char ++ 第二个t里面的--
public class Solution { public boolean isAnagram(String s, String t) { int [] counts = new int [ 26 ]; if ( s.length() != t.length () ) return false; for ( int i = 0; i < s.length(); i ++ ){ counts [ s.charAt( i ) - 'a' ] ++; counts [ t.charAt( i ) - 'a' ] -- ; } for ( int i = 0; i < 26; i ++ ){ if ( counts[ i ] != 0 ) return false; } return true; }}
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