【题目19】一些简单的算法练习题(1)

1. 比较两个字符串，用O(n)时间和恒量空间。

 

源代码：

 #include <stdio.h>#include <assert.h>int compare(const char *s1, const char *s2){assert(s1 != NULL && s2 != NULL);const char *p = s1;const char *q = s2;while( *p != '/0' && *q != '/0'){if(*p++ == *q++)continue;else if(*p++ > *q++)return 1;elsereturn -1;}if(*p == '/0' && *q != '/0')return -1;if(*p != '/0' && *q == '/0')return 1;return 0;}int main(){char* s1 = "hello";char* s2 = "world";int ret = compare(s1,s2);printf("compare /"%s/" and /"%s/"/n",s1,s2);if(ret == 0)printf("s1 = s2/n");else if(ret == -1)printf("s1 < s2/n");else printf("s1 > s2/n");s1 = "good";s2 = "good";printf("compare /"%s/" and /"%s/"/n",s1,s2);ret = compare(s1,s2);if(ret == 0)printf("s1 = s2/n");else if(ret == -1)printf("s1 < s2/n");else printf("s1 > s2/n");s1 = "aaabbb";s2 = "aaab";ret = compare(s1,s2);printf("compare /"%s/" and /"%s/"/n",s1,s2);if(ret == 0)printf("s1 = s2/n");else if(ret == -1)printf("s1 < s2/n");else printf("s1 > s2/n"); return 0;}

 2. 假设你有一个用1001个整数组成的数组，这些整数是任意排列的，但是你知道所有的整数都在

1到1000(包括1000)之间。此外，除一个数字出现两次外，其他所有数字只出现一次。假设你只

能对这个数组做一次处理，用一种算法找出重复的那个数字。如果你在运算中使用了辅助的存储方

式，那么你能找到不用这种方式的算法吗？

#include <stdio.h>int FindShowTwice(int *a,int len){bool *b = new bool[len-1];int i;for(i = 0; i < len-1; i++)b[i] = false;for(i = 0; i < len; i++){if(b[a[i] - 1] == true)return a[i];if(b[a[i] - 1] == false)b[a[i] - 1] = true;}delete[] b;return -1;}int main(){int a[1001];for(int i = 0; i < 1001; i++)a[i] = i+1;a[1000] = 15;printf("%d /n",FindShowTwice(a,1001));return 0;}

3.给定两个排好序的数组，怎样高效得判断这两个数组中存在相同的数字

#include <stdio.h>bool FindSameNum(int*a,int lenA, int* b, int lenB){int i = 0, j = 0;while(true){if(i == lenA && j == lenB)return false;if(a[i] == b[j])break;if(a[i] > b[j])j++;if(a[i] < b[j])i++;}return true;}int main(){int a[] = {2,3,5,6,7,8,32,43};int b[] = {4,12,24,32,44,55};if(FindSameNum(a,sizeof(a)/sizeof(int),b,sizeof(b)/sizeof(int)))printf("Have the same number./n");elseprintf("Don't have the same number./n");return 0;}

4. 删除字符串中指定的某个字符

#include <stdio.h>void DelSomeChar(char* str, int len, char c){int i = 0,j = 0;for(i = 0; i < len; i++)if(str[i] != c)str[j++] = str[i];str[j] = '/0';}int main(){char s[] = "abafdfafsdfasdfasdfaa";int len = sizeof(s)/sizeof(char);DelSomeChar(s,len,'a'); printf("%s/n",s);return 0;}

5. 删除字符串中特定的多个字符（>=1)

解法转载自：http://zhedahht.blog.163.com/blog/static/25411174200801931426484/

#include <stdio.h>#include <stdlib.h>#include <string.h>void DelMultiChars(char *pStrSrc, char* pStrDel){if(NULL == pStrSrc || NULL == pStrDel)return;const unsigned int nTableSize = 256;int hashTable[nTableSize];memset(hashTable,0,sizeof(hashTable));const char*pTemp = pStrDel;while('/0' != *pTemp){hashTable[*pTemp] = 1;++pTemp;}char* pSlow = pStrSrc;char* pFast = pStrSrc;while('/0' != *pFast){if(1 != hashTable[*pFast]){*pSlow = *pFast;++pSlow;}++pFast;}*pSlow = '/0';}int main(){char src[] = "hello,world";char del[] = "wl";printf("%s/n",src);DelMultiChars(src,del);printf("%s/n",src);return 0;}

6. 请定义一个宏，比较两个数a、b的大小，不能使用大于、小于、if语

用一个简单的数学知识来解答。

#define MAX(a,b) ((a+b)+ abs(a-b))/2