百度笔试题及答案

 

baidu笔试

一串首尾相连的珠子（m个），有N种颜色（N < 10），设计一个算法，取出其中一段，要求包含所有N中颜色，并使长度最短

 

 

将珠子映射为int,颜色映射为不同的值。

先定义一个简单的环状链表模板类RingList用于存储珠子，然后定义一个findMinFullRope函数用于计算。

 

#pragma oncetemplate <class T>class RingList{typedef unsigned int al_size;typedef int al_index;struct Node{Node(const T& t, Node* next = nullptr){_ = t;_next = next;}T _;Node* _next;};public:RingList(const T* values, al_size nSize){if (nSize > 0){_ = new Node(values[0]);Node* pNext = _;Node* p = nullptr;while(--nSize){p = new Node(values[nSize], pNext);pNext = p;}if (p){_->_next = p;}}}~RingList(){if (_ == nullptr) return;Node* p = _;do{Node* next = p->_next;delete p;p = next;}while(p != _);}T& operator [](al_index nIndex){if (_ == nullptr){T t;return t;}Node* p = _;while(--nIndex >= 0){p = p->_next;}return p->_;}private:Node* _;}; /*baidu笔试一串首尾相连的珠子（m个），有N种颜色（N < 10），设计一个算法，取出其中一段，要求包含所有N中颜色，并使长度最短*/#include "RingList.h"#include <set>bool findMinFullRope(int* balls, int m, int N, int& subFirst, int& subLast){if (m < 1 || N < 1) return false;if (m < N) return false;if (N == 1){subFirst = 0;subLast = 0;return true;}//m >= N > 1RingList<int> ring(balls, m);int start = 0;int end = 0;//找到第一个子串std::set<int> subRope;while (subRope.size() < N){if (end == m - 1){return false; //整串都不满足要求}subRope.insert(ring[++end]);}subFirst = start;subLast = end;int k = end; //第一次满足要求的子串的endint ballCount = end - start + 1; //子串球数int newCount = 0;while (&ring[end] != &ring[k - 1]) //end向后遍历，直到第一次end的前两位{//子串已经达到最少球数,无需再找if (ballCount == N){break;}//去掉子串前面的冗余的球do{subRope.clear();++start;for (int i = start; i <= end; ++i){subRope.insert(ring[i]);}}while (subRope.size() == N);--start;//去掉冗余球后子串的球数newCount = end - start + 1;//如果新满足条件的子串的球数较少,则更新球数和结果if (newCount < ballCount){ballCount = newCount;subFirst = start;subLast = end;}++end;//end向后遍历}//环状索引,可能超过球数,故取余subFirst %= m;subLast %= m;return true;}#include <stdio.h>#include <TCHAR.h>#include <iostream>extern bool findMinFullRope(int* balls, int m, int N, int& subFirst, int& subLast);int _tmain(int argc, _TCHAR* argv[]){int balls[] = {1,1,3,2,4,6,5,1,8,9,3,2,1,4,6,7,2,4,7,9,2,1,5,6,4};int m = sizeof(balls) / sizeof(int);std::cout<<"balls: ";for (int i = 0; i < m; ++i){std::cout<<balls[i]<<" ";}std::cout<<std::endl;int N = 2;int subFirst, subLast;if (findMinFullRope(balls, m, N, subFirst, subLast)){std::cout<<"startIndex : "<<subFirst<<"/tstart ball: "<<balls[subFirst]<<"/nendIndex: "<<subLast<<"/tend ball: "<<balls[subLast]<<std::endl;}else{std::cout<<"no valid result."<<std::endl;}getchar();return 0;}