Leetcode: Find the Difference

本文介绍了一种算法,该算法能够在两个字符串中找出被添加的一个字母。通过计算两个字符串中字符的ASCII值总和并求差值,进而得出新增的字母。此方法的时间复杂度为O(N),空间复杂度为O(1)。

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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.

O(N)time, O(1) space

 1 public class Solution {
 2     public char findTheDifference(String s, String t) {
 3         // Initialize variables to store sum of ASCII codes for 
 4         // each string
 5         int charCodeS = 0, charCodeT = 0;
 6         // Iterate through both strings and char codes
 7         for (int i = 0; i < s.length(); ++i) charCodeS += (int)s.charAt(i);
 8         for (int i = 0; i < t.length(); ++i) charCodeT += (int)t.charAt(i);
 9         // Return the difference between 2 strings as char
10         return (char)(charCodeT - charCodeS);
11     }
12 }

 

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