Commit e525fa9
suke
| Original file line number | Diff line number | Diff line change | |
|---|---|---|---|
@@ -0,0 +1,36 @@ | |||
| 1 | + class Solution { | ||
| 2 | + public boolean isAnagram(String s, String t) { | ||
| 3 | + HashMap<Character, Integer> stat = new HashMap<>(); | ||
| 4 | + for(int i = 0 ;i < s.length();i ++){ | ||
| 5 | + char c = s.charAt(i); | ||
| 6 | + Integer count = stat.get(c); | ||
| 7 | + if(count == null || count == 0){ | ||
| 8 | + stat.put(c,1); | ||
| 9 | + continue; | ||
| 10 | + } | ||
| 11 | + stat.put(c,count + 1); | ||
| 12 | + } | ||
| 13 | + for(int i = 0; i < t.length(); i ++){ | ||
| 14 | + char c = t.charAt(i); | ||
| 15 | + Integer count = stat.get(c); | ||
| 16 | + if(count == null){ | ||
| 17 | + return false; | ||
| 18 | + } | ||
| 19 | + if(count == 1){ | ||
| 20 | + stat.remove(c); | ||
| 21 | + continue; | ||
| 22 | + } | ||
| 23 | + if(count > 1){ | ||
| 24 | + stat.put(c,count - 1); | ||
| 25 | + } | ||
| 26 | + } | ||
| 27 | + return stat.isEmpty(); | ||
| 28 | + } | ||
| 29 | + public boolean isAnagram(String s, String t) { | ||
| 30 | + int[] alphabet = new int[26]; | ||
| 31 | + for (int i = 0; i < s.length(); i++) alphabet[s.charAt(i) - 'a']++; | ||
| 32 | + for (int i = 0; i < t.length(); i++) alphabet[t.charAt(i) - 'a']--; | ||
| 33 | + for (int i : alphabet) if (i != 0) return false; | ||
| 34 | + return true; | ||
| 35 | + } | ||
| 36 | + } | ||
| Original file line number | Diff line number | Diff line change | |
|---|---|---|---|
@@ -1 +1,4 @@ | |||
| 1 | - # 学习笔记 | ||
| 1 | + # 学习笔记 | ||
| 2 | + | ||
| 3 | + ## 242 | ||
| 4 | + - 有时候单纯的看时间复杂度，并不能很好的度量一个算法的性能，尤其是在不同数据量的情况下，第一种方法时间复杂度是2n，第二种是3n，但是整体下来，第二种方法却更快一些，使用数组代替hashmap能够更快的随机访问以外，尽量少的使用对象包装器，也会提高性能 | ||
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