剑指offer中一些有关数组和字符串操作的代码实现和扩充

本文主要介绍剑指offer中一些有关数组和字符串操作的C语言代码实现和扩充。

#include "vector"#include "stdio.h"#include "stdlib.h"#include <algorithm>#include <iostream>#include<math.h>using namespace std;//******************--求指针数组的长度--*********************int main31(){char *p[]={"zsxdzx","sdf","ZAcds"};//int s = sizeof(p)/sizeof(p[0]);int s = sizeof(p)/sizeof(*p);printf("%d\n",s);printf("%c\n",p[1][0]);return 0;}//******************--快速排序--*********************void Swap(int *num1, int *num2){int temp = *num1;*num1 = *num2;*num2 = temp;}int Partition(int data[], int length, int start, int end){if(data == NULL || length <=0 || start <0 || end >= length)return -1;int index = (rand() % (end-start+1))+ start;Swap(&data[index],&data[end]);int small = start-1;for (index=start;index<end;index++){if(data[index]<data[end]){++small;if(index != small)Swap(&data[index],&data[small]);}}++small;Swap(&data[index],&data[small]);return small;}void QuickSort(int data[], int length, int start, int end){if(data == NULL || length <=0 || start <0 || end >= length)return;int index = Partition(data,length,start,end);if(index>start)QuickSort(data,length,start,index-1);if(index<end)QuickSort(data,length,index+1,end);}int main32(){int number[] = {3,6,78,9,1,2,4,12,3};int startNum = 0;int endLen = sizeof(number)/sizeof(number[0]);QuickSort(number, endLen, startNum, endLen-1);//int num1=0, num2 = 10;//Swap(&num1,&num2);//printf("%d,%d\n",num1,num2);for (int i=0; i<endLen;i++)printf("%d\t",number[i]);return 0;}//******************--年龄排序--*********************void SortAge(int ages[], int length){if(ages == NULL || length <=0)return;const int oldAge = 99;int timeOfAge[oldAge+1];for(int i=0;i<=oldAge;i++)timeOfAge[i] = 0;for(int i=0; i<length; i++){int age = ages[i];if(age <=0 || age >oldAge)throw std::exception("input age wrong!");timeOfAge[age]++;}int index=0;for (int i=0;i<=oldAge;i++){for(int j=0;j<timeOfAge[i];j++){ages[index] = i;index++;}}}int main33(){int age[] = {15,16,72,25,21,23,22,21,22,21,23,25,29,30,26};int length = sizeof(age)/sizeof(age[0]);SortAge(age,length);for (int i=0; i<length;i++)printf("%d, ",age[i]);return 0;}//******************--字符串中去除重复的字符--改变原字符顺序*********************char* SortString(char *str, int length){if(str == NULL || length <= 0)return NULL;const int numbers = 52;int mychar[numbers+1];for(int i=0;i<=numbers;i++)mychar[i] = 0;for(int i=0; i<length-1; i++){int age = str[i]-'A';if(age <0 || (age >26 && age<32) || age > 58){printf("input age wrong!");return NULL;}if (age >=32)age -= 6;mychar[age]++;}int index = 0;for (int i=0;i<=numbers;i++){if (mychar[i] == 1){if (i<26)str[index] = i+'A';else if(i<52)str[index] = i+'A'+6;index++;}}for(int i=index;i<length;i++)str[i] = '\0';return str;}int main341(){char str[] = "abcabsdasdaAAA";int length = sizeof(str)/sizeof(str[0]);char* mystr = SortString(str,length);if(mystr == NULL)return -1;int length2 = strlen(mystr);for (int i=0; i<length2;i++)printf("%c",mystr[i]);printf("\n");return 0;}

//******************--字符串中去除重复的字符--未改变原字符顺序*********************char* SortString2(char *str, int length){  if(str == NULL || length <= 0)  return NULL;  const int numbers = 26;  int mychar[numbers];  char* tmp = (char*)malloc(sizeof(char)*length);memset(tmp,0,length);for(int i=0;i<numbers;i++)  mychar[i] = 0;  for(int i=0; i<length-1; i++){  tmp[i] = str[i];int s = str[i]-'a';  if(s <0 || s > 25){  printf("input age wrong!");  return NULL;  }  mychar[s]++;  }  int index = 0; for (int i=0;i<length;i++)  {  if(mychar[str[i]-'a'] == 1){str[index] = tmp[i];index++;}}  for(int i=index;i<length;i++)  str[i] = '\0';  return str;  }  int main342()  {  char str[] = "abcabsdasdathfgr";  int length = sizeof(str)/sizeof(str[0]);  char* mystr = SortString2(str,length);  if(mystr == NULL)  return -1;  int length2 = strlen(mystr);  for (int i=0; i<length2;i++)  printf("%c",mystr[i]);  printf("\n");  return 0;  }
//******************--求旋转数组中的最小值--*********************int minNum(int *numbers, int length){if(numbers == NULL || length <=0)return -1;int res = numbers[0];for(int i=1;i<length;i++){if(res > numbers[i])res = numbers[i];}return res;}int Min(int *numbers, int length){if(numbers == NULL || length <=0)return -1;int index1 = 0;int index2 = length-1;int mid = index1;while(numbers[index1]>=numbers[index2]){if(index2-index1 == 1){mid = index2;break;}mid = (index1+index2)/2;if(numbers[index1]==numbers[mid] && numbers[index2] == numbers[mid])return minNum(numbers,length);if(numbers[mid]>numbers[index1])index1 = mid;else if(numbers[mid] <= numbers[index2])index2 = mid;}return numbers[mid];}int main36(){int numbers[] = {1,1,1,0,1};int length = sizeof(numbers)/sizeof(numbers[0]);int res = 0;res = Min(numbers, length);printf("%d\n",res);return 0;}//******************--动态规划问题---剪绳子问题--*********************int maxProductAfterCutting_solution(int num){if(num<2)return 0;if(num==2)return 1;if(num==3)return 2;int* product = new int[num+1];product[0] = 0;product[1] = 1;product[2] = 2;product[3] = 3;int max = 0;for (int i=4;i<=num;i++){max = 0;for (int j=1;j<=i/2;j++){int prod = product[j]*product[i-j];if(prod>max)max = prod;//product[i] = max;}product[i] = max;}max = product[num];return max;}//******************-贪心算法---剪绳子问题--*********************int maxProductAfterCutting_solution2(int num){if(num<2)return 0;if(num==2)return 1;if(num==3)return 2;int timeOf3 = num/3;if (num%3 == 1)timeOf3--;int timeOf2 = (num-3*timeOf3)/2;return (int)(pow((double)3,timeOf3)) * (int)(pow((double)2,timeOf2));}int main35(){int num = 10;int res = 0;res = maxProductAfterCutting_solution(num);printf("%d\n",res);int res2 = maxProductAfterCutting_solution2(num);printf("%d\n",res2);return 0;}//******************--求数字对应的二进制中1的个数--*********************//方法一int NumberOf1(int num){int count = 0;unsigned int flag = 1;while(flag){if(num & flag)count ++;flag = flag << 1;}return count;}//方法二int NumberOf1_low(int num){int count = 0;while(num){count ++;num = (num-1)#}return count;}//******************--判断一数字是否为2的次方--*********************bool PowOf2(int num){if(num&(num-1))return false;elsereturn true;}//******************-求一个数到另一个数改变的步数-*********************int changeStride(int num1, int num2){//求一个数到另一个数改变的步数，先将两个数异或，再求结果中1的个数return NumberOf1(num1^num2);}int main37(){int number = 8;int res = 0;res = NumberOf1(number);printf("%d\n",res);int res2 = NumberOf1_low(number);printf("%d\n",res2);bool res3 = PowOf2(number);printf("%d\n",res3);int res4 = changeStride(10,13);printf("%d\n",res4);return 0;}//******************-数值的整数次方-*********************//方法一double PowerWithUnsignedExponent(double base, int exponent){double res = 1.0;for (int i=0;i<exponent;i++)res *= base;return res;}//方法二double PowerWithUnsignedExponent2(double base, int exponent){if(exponent==0)return 1.0;if(exponent == 1)return base;double res = PowerWithUnsignedExponent2(base, exponent>>1);res *= res;if((exponent & 0x1) == 1)res *= base;return res;}double Power(double base, int exponent){double res = 0;if(abs(base-0.0)<1e-6 && exponent<0)return 0.0;unsigned int absExponent = (unsigned int)exponent;if(exponent <0)absExponent = (unsigned int)(-exponent);//res = PowerWithUnsignedExponent(base,absExponent);res = PowerWithUnsignedExponent2(base,absExponent);if(exponent<0)res = 1.0/res;return res;}int main38(){double base = 2;int exponent = -2;double res = 0;res = Power(base, exponent);printf("%lf\n",res);return 0;}//******************-从1打印到最大的n位数-*********************void PrintNumber(char* number){if(number == NULL)return;bool flag = false;int len = strlen(number);for(int i=0;i<len;i++){if(!flag &&number[i] != '0')flag = true;if(flag)printf("%c",number[i]);}printf(", ");}//方法一bool Increment(char* number){if(number==NULL)return false;bool nFlow = false;int nTakeOver = 0;int len = strlen(number);for (int i=len-1;i>=0;i--){int nSum = number[i] - '0' + nTakeOver;if(i == len-1)nSum++;if(nSum>=10){if(i==0){nFlow = true;//return true;}else{nSum -= 10;nTakeOver = 1;number[i] = '0' + nSum;}}else{number[i] = '0' + nSum;break;}}return nFlow;}void Print1ToMaxOfN(int n){if(n<=0)return;char *number = new char[n+1];memset(number,'0',n);number[n] = '\0';while(!Increment(number)){PrintNumber(number);}delete []number;}//方法二：递归void Print1ToMaxOfN2_Recu(char* number, int n, int index){if(index == n-1){PrintNumber(number);return;}for(int i=0;i<10;i++){number[index+1] = i + '0';Print1ToMaxOfN2_Recu(number,n,index+1);}}void Print1ToMaxOfN2(int n){if(n<=0)return;char* number = new char[n+1];number[n] = '\0';for(int i=0;i<10;i++){number[0] = i + '0';Print1ToMaxOfN2_Recu(number,n,0);}delete []number;}int main39(){Print1ToMaxOfN(2);printf("\n");Print1ToMaxOfN2(2);return 0;}int GetNumberOfK2(int *a,int start, int end, int i){if(start == end)return a[start];int index = Partition(a,end+1,start,end);int k = index - start;if(i == k)return a[index];else if(i<k)return GetNumberOfK2(a, start, index-1, i);elsereturn GetNumberOfK2(a, index+1, end, i-k);}int GetNumberOfK(int *a,int start, int end, int k){if(start == end)return a[start];int index = Partition(a,end+1,start,end);while(index != k-1){if(index>k-1)index = Partition(a, end+1,start, index-1);elseindex = Partition(a, end+1,index+1, end);}return a[index];}int main390(){int a[] = {1,4,78,1,2,46,34,24};int ret = GetNumberOfK2(a,0,7,5);printf("%d\n",ret);return 0;}