You are given two strings of the same length 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.
Constraints:
1 <= s.length <= 5 * 104s.length == t.lengthsandtconsist of lowercase English letters only.
这道题给了两个字符串s和t，说是可以替换t中的字符，问最少替换多少个字符可以使得其与s是变位词。所谓的变位词，就是两个单词中字符的种类和个数均相同，就是字符的顺序不同而已。之前的题目中也做过不少关于变位词的题目，比如 Valid Anagram，Group Anagrams，Find Anagram Mappings，Find All Anagrams in a String 等等。这类题目的核心就是统计字符的出现次数，这道题也不例外，这里使用一个 HashMap 来统计字符串s中每个字符的出现次数。然后遍历字符串t，对于每个遍历到的字符，将 HashMap 中该字符的映射值自减1，这样操作之后映射值就可正可负，还可能为0。当某个映射值为正数的时候，则说明该字符在s中的数量多，若为负数，则说明该字符在t中的数量多，若为0，则说明该字符在s和t中的个数一样多。由于字符串s和t的长度相同，则正数的映射值累加和一定等于负数映射值的累加和，而且只要将所有的正数的映射字符替换成负数的映射字符，则s和t就会变成变位词，且替换次数最少，参见代码如下：
class Solution {
public:
int minSteps(string s, string t) {
int res = 0;
unordered_map<char, int> charCnt;
for (char c : s) ++charCnt[c];
for (char c : t) --charCnt[c];
for (auto a : charCnt) {
if (a.second > 0) res += abs(a.second);
}
return res;
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/1347
类似题目：
Determine if Two Strings Are Close
Minimum Number of Steps to Make Two Strings Anagram II
参考资料：
https://leetcode.com/problems/minimum-number-of-steps-to-make-two-strings-anagram/
LeetCode All in One 题目讲解汇总(持续更新中...)
标签：字符,映射,Make,Two,leetcode,anagram,Anagram,Strings,make From： https://www.cnblogs.com/grandyang/p/17459335.html