1、题目
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.
please:
Input:
s = "a"
t = "aa"
Output:
a
2、代码实现
public class Solution { public static char findTheDifference(String s, String t) { if (s == null || t.length() == 0) return t.charAt(0); if (t == null || t.length() == 0) return s.charAt(0); if (s == null && t == null) return 0; int[] a = new int[30]; char[] tChars = t.toCharArray(); char[] sChars = s.toCharArray(); int sLength = s.length(); int tLength = t.length(); if (sLength > tLength) { for (int i = 0; i < sChars.length; i++) { if (a[sChars[i] - 97] != 0) a[sChars[i] - 97] = ++(a[sChars[i] - 97]); else a[sChars[i] - 97] = 2; } for (int i = 0; i < tChars.length; i++) { a[tChars[i] - 97] = --(a[tChars[i] - 97]); } } else { for (int i = 0; i < tChars.length; i++) { if (a[tChars[i] - 97] != 0) a[tChars[i] - 97] = ++(a[tChars[i] - 97]); else a[tChars[i] - 97] = 2; } for (int i = 0; i < sChars.length; i++) { a[sChars[i] - 97] = --(a[sChars[i] - 97]); } } for (int i = 0; i < 30; i ++) { if (a[i] >= 2) { return (char) (i + 97); } } return 0; } }
3、总结
看到2个字符串对比，我么可以先转化为字符数组，下表从A - 65 活着 a - 95 开始，也就是从下表0开始，然后要注意2个字符串里面可能包含同样的元素有几个的情况，相同就往上加，另外一个就减，但是他们最多相差1个字符，所以，我们可可以肯定，比我们一开始设置的大，也就是3，然后如果没有重复的数据，那么一样的就为1，肯定有一个为2.
