一个腾讯的笔试题，比较有意思

 

 

腾讯的一道面试题，感觉值得收藏和讨论

 

http://topic.csdn.net/u/20110110/17/dcdb6234-842b-48be-a286-6ffb5b2534f4.html

 

 

假设两个字符串中所含有的字符和个数都相同我们就叫这两个字符串匹配，比如：abcda和adabc,由于出现的字符个数都是相同，只是顺序不同，所以这两个字符串是匹配的。。要求高效
现在需要你来实现下面的函数：
boolen Is_Mach(char *str1,char *str2)

 

方法1 

#include<stdio.h> #include<stdlib.h> #include<assert.h> #include<string.h> int Is_Mach(const char *src1, const char *src2) { unsigned int all_1[256] = {0}; unsigned int all_2[256] = {0}; const char *p = src1; assert (src1 != NULL && src2 != NULL); if (strlen(src1) != strlen(src2)) return 0; while (*p) { all_1[*p]++; p++; } while (*src2) { all_2[*src2]++; src2++; } while (*src1) { if (all_1[*src1] != all_2[*src1]) return 0; src1++; } return 1; } int main() { char src1[] = "abcdefdjaidjadjapodjasojfidofhdsufhdshfdshgufdghufdghudfghudhgid"; char src2[] = "japodjasojfidofhdsuabcdefdjaidjadfhdshfhufdghudfgdshgufdghudhgid"; printf("%d/n", Is_Mach(src1, src2)); system("pause"); return 0; }  

 

 

方法2 

bool Is_Mach(char *str1,char *str2) { const int MAX = 255; int histogram1[MAX]; int histogram2[MAX]; memset(histogram1, 0, MAX); memset(histogram2, 0, MAX); int len1 = strlen(str1); int len2 = strlen(str2); if (len2 != len1) return false; while(*str1) histogram1[ *str1++ ]++; while(*str2) histogram2[ *str2++ ]++; bool flag=true; for (int i=0; i<MAX; i++) flag &= ( histogram1[i]==histogram2[i] ); return flag; } 

 

3

bool Is_Mach(char *str1, char *str2) { bool isMach = strlen(str1) == strlen(str2); if (isMach) { int count[256] = {1}; for(char *p = str1; *p; ++count[*p++]) ; for(char *p = str2; *p; ++p) { if(--count[*p] == 0) { isMach = false; break; } } } return isMach; } 

 

4

bool Is_Mach(char *str1, char *str2) { unsigned int isMach = strlen(str1) - strlen(str2); if (isMach == 0) { unsigned int count[256] = {0}; for(char *p = str1; *p; ++count[*p++]) ; for(char *p = str2; *p; isMach |= --count[*p++]) ; isMach >>= sizeof(unsigned int) * 8 - 1; } return isMach == 0; }