題目
Given two equal-size strings s and t. In one step you can choose any character of t and replace it with another character.
Return the minimum number of steps to make t an anagram of s.
An Anagram of a string is a string that contains the same characters with a different (or the same) ordering.
Example 1:
Input: s = “bab”, t = “aba”
Output: 1
Explanation: Replace the first ‘a’ in t with b, t = “bba” which is anagram of s.
Example 2:
Input: s = “leetcode”, t = “practice”
Output: 5
Explanation: Replace ‘p’, ‘r’, ‘a’, ‘i’ and ‘c’ from t with proper characters to make t anagram of s.
Example 3:
Input: s = “anagram”, t = “mangaar”
Output: 0
Explanation: “anagram” and “mangaar” are anagrams.
Example 4:
Input: s = “xxyyzz”, t = “xxyyzz”
Output: 0
Example 5:
Input: s = “friend”, t = “family”
Output: 4
Constraints:
1 <= s.length <= 50000
s.length == t.length
s and t contain lower-case English letters only.
分析
題意：t和s具有相同的字母（不考慮順序），t就是s的Anagram。
給出t需要變幾個字母才能成爲s的Anagram。
算法：
一共就26個字母，只需要構造一個26位的數組，數組的下標0~25分別表示a~z，然後遍歷s，讓數組對應下標位置++。
接着遍歷t讓數組對應所在位有值的部分–。
最後對數組元素求和即可。
解答
class Solution {
public int minSteps(String s, String t) {
int[] flags = new int[26];
for(char c:s.toCharArray()){
flags[c-'a']++;
}
for(char c:t.toCharArray()){
if(flags[c-'a']>0){
flags[c-'a']--;
}
}
int sum=0;
for(int i=0;i<26;++i){
sum+=flags[i];
}
return sum;
}
}