题目:写一个函数, 如 Foo(const char *str), 打印出 str 的全排列,
如 abc 的全排列: abc, acb, bca, dac, cab, cba
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 | #include “stdafx.h” #include <iostream> #include <iterator> #include <string> #include <vector> #include <algorithm> using namespace ::std; void Foo(const char *str); //stl void Foo2(const char *str); //recursion void foo2( char *str,int position,int strlen ,int &count); void Foo3(const char *str); //loop int main(int argc, char *argv[]) { unsigned long start = ::GetTickCount(); Foo(”abcdefg”); unsigned long end = ::GetTickCount(); unsigned long worktime1 = end - start; start = ::GetTickCount(); Foo2(”abcdefg”); end = ::GetTickCount(); unsigned long worktime2 = end - start; start = ::GetTickCount(); Foo3(”abcdefg”); end = ::GetTickCount(); unsigned long worktime3 = end - start; cout< <”use stl time = “<<worktime1 <<endl; cout<<”use recursion time = “<<worktime2 <<endl; cout<<”use loop time = “<<worktime3 <<endl; system(”pause”); return 0; } void Foo(const char *str) { int n=strlen(str),count=0; vector<char> a(n); for(int i=0; i<n ; i++) a[i]=str[i]; vector<char>::iterator start, end; ostream_iterator<char> outIt(cout, “”); start = a.begin(); end = a.end(); do { count++; copy(start, end, outIt) ; cout < < endl ; } while(next_permutation(start, end)); cout<<count<<endl; } void Foo2(const char *str) { int n=strlen(str),count=0; char *mystr = (char *)malloc(n*sizeof(char)); memcpy(mystr , str ,n); foo2(mystr , 0 , n ,count); cout<<count<<endl; free(mystr); } void foo2(char *str,int position,int strlen ,int &count) { if(position == strlen) { cout<<str<<endl; count ++; } else { for(int i = position; i < strlen; i++) { swap(*(str + i) , *(str + position)); foo2(str,position+1,strlen,count); swap(*(str + i) , *(str + position)); } } } void Foo3(const char *str) { int len = strlen(str); int * range = new int[len]; int cont, i, j, counter = 0; memset(range, 0, len*sizeof(int)); while(1) { for(i=0; i<len; i++) { if(++range[i]>=len) range[i] = 0; else break; } if (i==len) { break; } for(i=0,cont=0; i<len -1; i++) { for(j=i+1; j<len; j++) { if(range[i] == range[j]) { cont=1; break; } } if (cont == 1) { break; } } if (cont) continue; for(i=0; i<len; i++) { printf(”%c”, str[range[i]]); } printf(”n”); ++counter; } delete [] range; cout<<counter<<endl; } |
据说是迅雷的一个笔试题目,这给了三种方法,第一种用库函数,当然这是一个玩笑.第二使用的递归,第三使用的是循环,第三个的思路比较特别,生成一个不重复的数列,一般的比较,递归是比较快的一种方法,其次是循环.这些代码都是来自csdn的讨论,我做了些修改和优化.还有,在stdafx.h中加入 #include "Windows.h",没有考虑有重复字符的问题
