leetcode解题: Find the Difference (389)

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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.

解法1：O(N) Time Complexity with O(1) Space

用Hash的思路，对短字符串建立Hash索引，然后长字符串里面每一个字符减一，最后如果发现出现负数次数的字符时则一定是多出来的那个字符。
C++ 中注意unordered_map的用法，unordered_map::operator[] 对于不存在的元素会进行插入操作并初始化对应的map数值。
Java取字符串中的字符的method是s.charAt(i)
C++

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class Solution {
public:
    char findTheDifference(string s, string t) {
        std::unordered_map<char, int> strmap;
        for (char c: s) ++strmap[c];
        for (char c: t) {
            --strmap[c];
            if (strmap[c] < 0) return c;
        }
        return 0;
    }
};

Java

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public class Solution {
    public char findTheDifference(String s, String t) {
       int[] map = new int[26];
       for (int i = 0; i < s.length(); i++) {
           map[s.charAt(i) - 'a']++;
           map[t.charAt(i) - 'a']--;
       }
       map[t.charAt(t.length() - 1) - 'a']--;
       char res = 'a';
       for (int i = 0; i < 26; i++) {
           if (map[i] < 0) {
               res += i;
               break;
           }
       }
       return res;
    }
}

解法2：O(N) Time Complexity with O(1) Space

找不同的题目可以尝试用抵消的思路。一个抵消的工具就是XOR，如果对每一个字符进行异或操作，由于相同的元素抵消，剩下的一定是不相同的那一个。
C++

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class Solution {
public:
    char findTheDifference(string s, string t) {
       char res = 0;
       for (char c: s) res ^= c;
       for (char c: s) res ^= c;
       return res;
    }
};