前言:继续分享,加油!
面试题44:
代码如下:
#include<iostream> #include<stdlib.h> using namespace std; int compare(const void *arg1,const void *arg2) { return *(int *)arg1 - *(int *)arg2; } bool IsContinuous(int *numbers,int len) { if(numbers==NULL || len<1) return false; qsort(numbers,len,sizeof(int),compare); int numberOfZero = 0; int numberOfGap = 0; //统计数组中的间隔数目 int small = numberOfZero; int big = small+1; while(big < len) { //两个数相等,有对子,不可能是顺子 if(numbers[small] == numbers[big]) return false; numberOfGap += numbers[big]-numbers[small]-1; small = big; ++big; } return (numberOfGap>numberOfZero)?false:true; } int main() { int data[]={1,2,3,4,9}; cout << boolalpha << IsContinuous(data,5) << endl; return 0; }
面试题45:
代码如下:
#include<iostream> #include<list> using namespace std; int LastRemaining(unsigned int n,unsigned int m) { if(n<1 || m<1) return -1; unsigned int i=0; //创建双向链表 list<int> numbers; for(i=0;i<n;++i) numbers.push_back(i); //使用迭代器 list<int>::iterator current = numbers.begin(); while(numbers.size() > 1) { //找到m位置 for(int i=1;i<m;++i) { current++; //将表尾指向表头 if(current == numbers.end()) current = numbers.begin(); } list<int>::iterator next = ++current; if(next == numbers.end()) next = numbers.begin(); --current; //删除节点 numbers.erase(current); current = next; } return *(current); } int main() { cout << "data: " << LastRemaining(5,3) << endl; return 0; }
面试题46:
这道题用是三种方法给大家演示:
方法一:构造函数
/* 解法一:利用构造函数 */ #include<iostream> using namespace std; class Temp{ private: static unsigned int N; static unsigned int Sum; public: //构造函数 Temp(){++N;Sum+=N;}; static void Reset(){N=0;Sum=0;} static unsigned int GetSum(){return Sum;} }; unsigned int Temp::N = 0; unsigned int Temp::Sum = 0; unsigned int Sum_Solution(unsigned int n) { Temp::Reset(); //会调用n次 Temp *a = new Temp[n]; delete []a; a=NULL; return Temp::GetSum(); } int main() { cout << "sum: " << Sum_Solution(100) << endl;; return 0; }
方法二:虚函数
/* 解法二 用虚函数 */ #include<iostream> using namespace std; class A; A*Array[2]; class A{ public: virtual unsigned int Sum(unsigned int n) { return 0; } }; class B:public A { public: virtual unsigned int Sum(unsigned int n) { //两次取反,非0的n转换为true return Array[!!n]->Sum(n-1) + n; } }; int Sum_Solution(int n){ A a; B b; Array[0] = &a; Array[1] = &b; int value = Array[1]->Sum(n); return value; } int main() { cout << "sum: " << Sum_Solution(3) << endl; return 0; }
方法三:利用函数指针
/* 利用函数指针 */ #include<iostream> using namespace std; typedef unsigned int (*fun)(unsigned int); unsigned int Solution_T(unsigned int n) { return 0; } unsigned int Sum_Solution(unsigned int n) { static fun f[2] = {Solution_T,Sum_Solution}; return n + f[!!n](n-1); } int main() { cout << "sum: " << Sum_Solution(100) << endl; return 0; }
面试题47:
代码如下:
#include<iostream> using namespace std; int Add(int num1,int num2) { int sum,carry; do{ //两个数先求异或,相同为0,相反为1 sum = num1 ^ num2; //相与然后左移一位 carry = (num1&num2)<<1; num1 = sum; num2 = carry; }while(num2 != 0); return num1; } int main() { cout << "num: " << Add(1,2) << endl; return 0; }
总结:经过很长一段时间才把这本书看完并码代码,收获很多,数据结构和算法之路才刚刚开始,学习之路还很长,希望对各位能有所帮助。继续加油...,梦想还是要用有的,万一见鬼了呢!
作者:
柳德维
-------------------------------------------
个性签名:独学而无友,则孤陋而寡闻。做一个灵魂有趣的人!
如果觉得这篇文章对你有小小的帮助的话,记得在右下角点个“推荐”哦,博主在此感谢!
万水千山总是情,打赏一分行不行,所以如果你心情还比较高兴,也是可以扫码打赏博主,哈哈哈(っ•̀ω•́)っ⁾⁾!