LeetCode 389. Find the Difference

Posted 2020-08-20

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.

本题最直观的思路为排序后对s和t逐位进行比较。故得以下代码：

 1 class Solution 
 2 {
 3 public:
 4     char findTheDifference(string s, string t) 
 5     {
 6         sort(s.begin(), s.end());
 7         sort(t.begin(), t.end());
 8         for(int i = 0; i < s.size(); i++)
 9         {
10             if(s[i] == t[i]) 
11             {
12                 continue;
13             }
14             else
15             {
16                 return t[i];
17             }
18         }
19         return t[t.size() - 1];
20     }
21 };

由于排序复杂度过高，于是查询别的解法发现，可以建立一个字母表，s出现一次字母表对应位置+1，t中出现一次对应位置-1。那么只在t中出现的字母的位置为-1。代码如下：

 1 public char findTheDifference(String s, String t) {
 2     if(s == null || s.length() == 0)
 3         return t.charAt(0);
 4     int[] letters = new int[26];
 5     for(int i = 0; i < s.length(); i++){
 6         int sPosition = s.charAt(i) - ‘a‘;
 7         int tPosition = t.charAt(i) - ‘a‘;
 8         letters[sPosition]++;
 9         letters[tPosition]--;
10     }
11     int tPosition = t.charAt(t.length()-1) - ‘a‘;
12     letters[tPosition]--;
13     char res = ‘a‘;
14     for(int i = 0; i < letters.length; i++){
15         if(letters[i] == -1){
16             res+= i;
17             break; 
18         }    
19     }
20     return res;
21 }