LeetCode之Find the Difference

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:
eExplanation:
'e' is the letter that was added.

please:

Input:

s = "a"

t = "aa"

Output:

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]);elsea[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]);elsea[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.

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LeetCode之Find the Difference

1、题目

2、代码实现

3、总结

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