Valid Anagram
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.
Note:
You may assume the string contains only lowercase alphabets. 题目的意思理解了anagram这个单词就理解了，这个词是“易位构词”的意思，其实就是指两个字符串的组成字母是否完全一致。 题目中给的两个例子也很明显。1
我的思路是先排序，然后挨个比较。自己实现的快排。
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public class Solution {
public boolean isAnagram(String s, String t) {
if(s == null || t == null){
return false;
}
if(s.length() != t.length()){
return false;
}
char[] schar = s.toCharArray();
char[] tchar = t.toCharArray();
quickSort(schar);
quickSort(tchar);
for(int i = 0; i < s.length();i++){
if(schar[i] != tchar[i]) return false;
}
return true;
}
public static void quickSort(char[] a){
int n = a.length;
quickSort(a,0,n-1);
}
private static void quickSort(char[] a,int lo,int hi){
if(lo >= hi) return;
int j = partition(a,lo,hi);
quickSort(a,lo,j-1);
quickSort(a,j+1,hi);
}
private static int partition(char[] a,int lo,int hi){
int i = lo;
int j = hi+1;
while(true){
while(less(a,++i,lo)) if(i == hi) break;
while(less(a,lo,--j)) if(j == lo) break;
if(i >= j) break;
swap(a,i,j);
}
swap(a,lo,j);
return j;
}
private static void swap(char[] a,int i,int j){
char tmp = a[i];
a[i] = a[j];
a[j] = tmp;
}
private static boolean less(char[] a, int i,int j){
return a[i] < a[j];
}
}
2 leetcode discuss
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public static boolean isValidAnagram(String s,String t){
if(s.length() != t.length()) return false;
int[] count = new int[26];
for(int i = 0; i < s.length(); i++){
count[s.charAt(i) - 'a']++;
count[t.charAt(i) - 'a']--;
}
for(int i = 0; i < count.length; i++){
if(count[i] != 0) return false;
}
return true;
}