一個騰訊的筆試題，比較有意思

 

 

騰訊的一道面試題，感覺值得收藏和討論

 

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

 

 

假設兩個字符串中所含有的字符和個數都相同我們就叫這兩個字符串匹配，比如：abcda和adabc,由於出現的字符個數都是相同，只是順序不同，所以這兩個字符串是匹配的。。要求高效
現在需要你來實現下面的函數：
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; }