1、題目

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: a

2、代碼實現

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、總結

看到2個字符串對比，我么可以先轉化為字符數組，下表從A - 65 活著 ?a - 95 ?開始，也就是從下表0開始，然后要注意2個字符串里面可能包含同樣的元素有幾個的情況，相同就往上加，另外一個就減，但是他們最多相差1個字符，所以，我們可可以肯定，比我們一開始設置的大，也就是3，然后如果沒有重復的數據，那么一樣的就為1，肯定有一個為2.

總結

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