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编写一个函数，作用是把一个char组成的字符串循环右移n个

Posted by zeaster in September 25th 2006

例如，”abcdef”循环右移2个就变为”efabcd”
popo实现了一个，不过需要O(n)的额外空间程序：void LoopMove(char * pStr,int steps){ int len = strlen(pStr); char *s = new char[len+1]; int count = 0; char *temp = pStr; while (*pStr) { s[(count+steps)%len] = *pStr; pStr++; count++; } s[len] = ‘\0′; strcpy(temp,s); delete []s;}
最近想出了个空间为O(1)，时间为O(n)的算法就是只通过数组指针移动和交换来实现
int LoopMove2(char *p, int i, int t, int s, int m){ int n = strlen(p); int [...]

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Google 面试第一题解法

Posted by zeaster in September 25th 2006

Solve this cryptic equation, realizing of course that values for M and E could be interchanged. No leading zeros are allowed.
WWWDOT - GOOGLE = DOTCOM
思路：穷举法，式子中共有9个不同的字母，数字0－9有10个，因此就从10个数字中任取9个，排成一列，这样分别对应了式子中的字母。然后判断当前取出的这9个数，是否满足条件。本程序中用check2 函数判断，另外需要考虑WE是可以互换，但不可以相同
答案：只有两组解777589 - 188103 = 589486777589 - 188106 = 589483
程序：Perm1.javapackage org.chao.perm;
public class Perm1 { public static void main(String[] args) { [...]

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编写一个函数，作用是把一个char组成的字符串循环右移n个

Google 面试第一题解法

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