Valid Password

最新推荐文章于 2022-12-20 15:56:05 发布

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最新推荐文章于 2022-12-20 15:56:05 发布

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原文链接：http://www.cnblogs.com/superbo/p/4111923.html

本文介绍了一种算法，用于判断在1-9数字键盘中有一个按键损坏的情况下，输入的密码是否有效。通过两个指针分别遍历预期密码和实际输入密码，并记录损坏按键，最后验证是否匹配。

Problem

In 1-9 keypad one key is not working. If someone enters a password then not working key will not be entered. You have given expected password and entered password. Check that entered password is valid or not. Ex: entered 164, expected 18684 (you need to take care as when u enter18684 and 164 only both will be taken as 164 input)

Solution

有几种情况需要考虑下，

1. 没有按到过坏的键

2. 1864， 1684. 坏的是8，坏的键又出现一次

2. expect有可能是186848888，要继续检查直到长度相等

 1 public static boolean validPassword(String expect, String actual) {
 2     char faultKey = '\0'; //initial fault key
 3     int i=0, j=0;    //two pointer for each string
 4     for(; i<actual.length() && j<expect.length(); j++) {
 5         if(actual.charAt(i) != expect.charAt(j)) {
 6             if(faultKey == '\0')    faultKey = expect.charAt(j);    //record fault key
 7             else if(faultKey != expect.charAt(j))    return false;    //found second fault key
 8             i--;
 9         }
10         i++;
11     }
12     
13     for(int k=0; k<expect.length() ;k++) {
14         if(faultKey == expect.charAt(k)) return false;            
15     }
16     //fault key may require press many times after that
17     while(j<expect.length() && expect.charAt(j)==faultKey)    j++;
18     
19     return (i==actual.length() && j==expect.length() && faultKey == '\0') ? true : false;
20 }