389. Find the Difference*

​​https://leetcode.com/problems/find-the-difference/​​

题目描述

Given two strings ​​s​​​ and ​​t​​ which consist of only lowercase letters.

String ​​t​​ is generated by random shuffling string s and then add one more letter at a random position.

Find the letter that was added in ​​t​​.

Example:

Input:
s = "abcd"
t = "abcde"

Output:
e

Explanation:
'e' is the letter that was added.

C++ 实现 1

使用哈希表统计 ​​s​​​ 和 ​​t​​ 中的个数.

class Solution {
public:
    char findTheDifference(string s, string t) {

        unordered_map<char, int> record;
        for (int i = 0; i < s.size(); ++i) {
            record[t[i]] ++;
            record[s[i]] --;
        }
        record[t[t.size() - 1]] ++;
        char res = '\0';
        for (auto &iter : record)
            if (iter.second != 0)
                res = iter.first;
        return res;
    }
};

C++ 实现 2

仍然是统计个数.

class Solution {
public:
    char findTheDifference(string s, string t) {
        vector<int> position(26, 0);
        for (int i = 0; i < s.size(); ++ i) {
            position[s[i] - 'a'] ++;
            position[t[i] - 'a'] --;
        }
        position[t.back() - 'a'] --;
        char res;
        for (int i = 0; i < position.size(); ++ i) {
            if (position[i] != 0) res = 'a' + i;
        }
        return res;
    }
};

请注意这段代码:

for (int i = 0; i < s.size(); ++ i) {
   position[s[i] - 'a'] ++;
    position[t[i] - 'a'] --;
}
position[t.back() - 'a'] --;

和 ​​C++ 实现 1​​​ 是不同的, 原因在于在 ​​for​​​ 循环中是先对 ​​s[i]​​​ 进行 ​​++​​​ 操作, 然后对 ​​t[i]​​​ 进行 ​​--​​​ 操作, 之后在 ​​for​​​ 循环外面, 处理 ​​t.back()​​​ 时, 要保证和 ​​for​​​ 循环的处理一致, 即也采用 ​​--​​​ 操作, 如果此时误写为 ​​++​​, 那么会引起 bug.

C++ 实现 3

使用异或操作.

class Solution {
public:
    char findTheDifference(string s, string t) {
        vector<int> position(26, 0);
        char res = '\0';
        for (int i = 0; i < s.size(); ++ i) {
            res ^= s[i];
        }
        for (int i = 0; i < t.size(); ++ i) {
            res ^= t[i];
        }
        return res;
    }
};