242. Valid Anagram
Posted 我的名字叫周周
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/* * 242. Valid Anagram * 2016-6-18 by Mingyang * 1是别人建的bucket2是自己见得bucket,原理一样,更简洁 * 特别注意人家的empty case一行代码就搞定,简单! */ public boolean isAnagram1(String s, String t) { if (s == null || t == null || s.length() != t.length()) return false; int[] map = new int[26]; for (int i = 0; i < s.length(); i++) map[s.charAt(i) - ‘a‘]++; for (int i = 0; i < t.length(); i++) if(--map[t.charAt(i) - ‘a‘] < 0) return false; return true; } public boolean isAnagram2(String s, String t) { if (s == null || t == null || s.length() != t.length()) return false; int[] map=new int[256]; Arrays.fill(map,0); for(int i=0;i<s.length();i++){ map[s.charAt(i)]++; } for(int i=0;i<t.length();i++){ map[t.charAt(i)]--; if(map[t.charAt(i)]<0) return false; } return true; } /* * Arrays.sort(int[]) in all Java standard library implementations that I know, * is an example of a comparison-based sort and thus must have worst-case complexity Ω(n log n). * In particular, Oracle Java 7 uses a dual-pivot quicksort variant for the integer overloads, * which actually has an Ω(n2) worst case. */ public static boolean isAnagram(String a, String b) { char[] aa = a.toCharArray(); char[] bb = b.toCharArray(); Arrays.sort(aa); String a1 = new String(aa); Arrays.sort(bb); String b1 = new String(bb); if (a1.equals(b1)) return true; else return false; }
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