LeetCode 1347. Minimum Number of Steps to Make Two Strings Anagram

[Med] LeetCode 1347. Minimum Number of Steps to Make Two Strings Anagram

链接: https://leetcode.com/problems/minimum-number-of-steps-to-make-two-strings-anagram/

题目描述：
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.

给你两个长度相等的字符串 s 和 t。每一个步骤中，你可以选择将 t 中的 任一字符 替换为 另一个字符。
返回使 t 成为 s 的字母异位词的最小步骤数。
字母异位词 指字母相同，但排列不同的字符串。

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.

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Tag: String
解题思路
检查从字符串A转到字符串B需要改变几个字母。首先保证了两个字符串长度相同。很简单，计算一下A中多了几个就可以了

解法一:

class Solution {
    public int minSteps(String s, String t) {
        int[] l1 = new int[26];
        int[] l2 = new int[26];
        for(int i=0; i<s.length(); i++){
            l1[s.charAt(i)-'a']++;
            l2[t.charAt(i)-'a']++;
        }
        
        for(int i=0; i<l1.length; i++){
            l1[i] -= l2[i];
        }
        int res =0;
        
        for(int i: l1) res+=i>0?i:0;
        return res;
    }
}

解法二:
可以进一步简化

class Solution {
    public int minSteps(String s, String t) {
        int[] list = new int[26];
       
        for(int i=0; i<s.length(); i++){
            list[s.charAt(i)-'a']++;
            list[t.charAt(i)-'a']--;
        }
        
        int res =0;
        
        for(int i: list) res+=i>0?i:0;
        
        return res;
    }
}