题目链接:https://leetcode.com/problems/valid-anagram/
题目:
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.
Follow up:
What if the inputs contain unicode characters? How would you adapt your solution to such case?
思路:
用HashMap存储s的字符和它出现次数的关系,遍历字符串t并减去hashmap中存储的相应的字符出现的次数,如果t中有字符不包含在hashmap中返回false,最后如果hashmap中value有不为0的,则返回false,否则返回true。本题因为是小写字母可以用26位数组表示对应字母出现的次数。实质也是hashmap的思想。
算法:
public boolean isAnagram(String s, String t) { int count[] = new int[26]; if (s.length() != t.length()) { return false; } if (s.equals(t)) { return true; } char ss[] = s.toCharArray(); char tt[] = t.toCharArray(); for (char tmp : ss) { count[tmp - 'a']++; } for (char tmp : tt) { count[tmp - 'a']--; } for (int i : count) { if (i != 0) { return false; } } return true; }