Leetcode: Find the Difference

找出字符串中新增字母的算法

weixin_30885111

于 2016-11-30 05:40:00 发布

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原文链接：http://www.cnblogs.com/EdwardLiu/p/6116154.html

本文介绍了一种算法，该算法通过计算两个字符串中字符的ASCII值总和来找出被随机插入的一个额外字母。这种方法的时间复杂度为O(N)，空间复杂度为O(1)。示例代码展示了如何实现这一算法。

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 }