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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.
(An anagram is a type of word play, the result of rearranging the letters of a word or phrase to produce a new word or phrase, using all the original letters exactly once; for example Torchwood can be rearranged into Doctor Who.)
Use hashmap to count each character‘s appearance time. Time complexity O(n), space cost O(n).
Note here, counts.put(tmp, count.get(tmp)--); is wrong!
1 public class Solution { 2 public boolean isAnagram(String s, String t) { 3 if (s == null || t == null) 4 return false; 5 if (s.length() != t.length()) 6 return false; 7 Map<Character, Integer> counts = new HashMap<Character, Integer>(); 8 int length = s.length(); 9 for (int i = 0; i < length; i++) { 10 char tmp = s.charAt(i); 11 if (counts.containsKey(tmp)) { 12 int count = (int)counts.get(tmp); 13 counts.put(tmp, count + 1); 14 } else { 15 counts.put(tmp, 1); 16 } 17 } 18 for (int i = 0; i < length; i++) { 19 char tmp = t.charAt(i); 20 if (counts.containsKey(tmp)) { 21 int count = (int)counts.get(tmp); 22 if (count <= 0) 23 return false; 24 else 25 counts.put(tmp, count - 1); 26 } else { 27 return false; 28 } 29 } 30 return true; 31 } 32 }
++x is called preincrement while x++ is called postincrement.
1 int x = 5, y = 5; 2 3 System.out.println(++x); // outputs 6 4 System.out.println(x); // outputs 6 5 6 System.out.println(y++); // outputs 5 7 System.out.println(y); // outputs 6
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原文地址：http://www.cnblogs.com/ireneyanglan/p/4809024.html
