每日一题(22) - 寻找数组中的最大值和最小值

题目来自编程之美

题目：寻找数组中的最大值和最小值

要求：要求比较次数最少

分析：使用分治法解决，详细见点击打开链接

代码：

#include <iostream>#include <assert.h>using namespace std;void FindMaxAndMin(int arr[],int nStart,int nEnd,int& nMax,int& nMin){assert(arr != NULL && nStart <= nEnd && nStart >= 0 && nEnd >= 0);int nLeftMax = 0;int nLeftMin = 0;int nRightMax = 0;int nRightMin = 0;if (nStart == nEnd - 1)//区间相差为1{if (arr[nStart] > arr[nEnd]){nMax = arr[nStart];nMin = arr[nEnd];}else{nMax = arr[nEnd];nMin = arr[nStart];}return;}if (nStart == nEnd)//区间相差为0{nMin = nMax = arr[nStart];return;}int nMid = (nStart + nEnd)/2;FindMaxAndMin(arr,nStart,nMid,nLeftMax,nLeftMin);FindMaxAndMin(arr,nMid + 1,nEnd,nRightMax,nRightMin);//分别比较两部分的最大值和最小值，求出整体的最大值和最小值nMax = max(nLeftMax,nRightMax);nMin = min(nLeftMin,nRightMin);}int main(){int nMin = 0;int nMax = 0;int arr[8] = {4,-3,5,-2,-1,2,6,-2};FindMaxAndMin(arr,0,7,nMax,nMin);cout<<"Max: "<<nMax<<endl;//6cout<<"Min: "<<nMin<<endl;//-3int arr1[8] = {4,-3,5,-11,4,-2,5,-2};FindMaxAndMin(arr1,0,7,nMax,nMin);cout<<"Max: "<<nMax<<endl;//5cout<<"Min: "<<nMin<<endl;//-11int arr2[8] = {-3,-5,-8,-2,-11,-12,-13,-1};FindMaxAndMin(arr2,0,7,nMax,nMin);cout<<"Max: "<<nMax<<endl;//-1cout<<"Min: "<<nMin<<endl;//-13system("pause");return 1;}

题目来自编程之美

题目：寻找数组中的最大值和次大值

要求：要求比较次数最少

分析：使用分治法解决，详细见点击打开链接

注意：如果最大值在序列中重复出现，无法找到次大值。

举例：如序列：4,-3,5,-11,4,-2,5,-2, 最大值5出现了两次，此时算法输出5和5,而不是5,4。

代码:

#include <iostream>#include <assert.h>using namespace std;void FindMaxAndSecMax(int arr[],int nStart,int nEnd,int& nMax,int& nSecMax){assert(arr != NULL && nStart <= nEnd && nStart >= 0 && nEnd >= 0);int nLeftMax = 0;int nLeftSecMax = 0;int nRightMax = 0;int nRightSecMax = 0;if (nStart == nEnd - 1)//区间相差为1{if (arr[nStart] > arr[nEnd]){nMax = arr[nStart];nSecMax = arr[nEnd];}else{nMax = arr[nEnd];nSecMax = arr[nStart];}return;}if (nStart == nEnd)//区间相差为0{nSecMax = nMax = arr[nStart];return;}int nMid = (nStart + nEnd)/2;FindMaxAndSecMax(arr,nStart,nMid,nLeftMax,nLeftSecMax);FindMaxAndSecMax(arr,nMid + 1,nEnd,nRightMax,nRightSecMax);//分别比较两部分的最大值和次大值，求出整体的最大值和次大值if (nLeftMax > nRightMax){nMax = nLeftMax;if (nLeftSecMax > nRightMax){nSecMax = nLeftSecMax;}else{nSecMax = nRightMax;}}else{nMax = nRightMax;if (nRightSecMax > nLeftMax){nSecMax = nRightSecMax;}else{nSecMax = nLeftMax;}}}int main(){int nMin = 0;int nMax = 0;int arr[8] = {4,-3,5,-2,-1,2,6,-2};FindMaxAndSecMax(arr,0,7,nMax,nMin);cout<<"Max: "<<nMax<<endl;//6cout<<"Min: "<<nMin<<endl;//5int arr1[8] = {4,-3,5,-11,4,-2,5,-2};FindMaxAndSecMax(arr1,0,7,nMax,nMin);cout<<"Max: "<<nMax<<endl;//5cout<<"Min: "<<nMin<<endl;//5int arr2[8] = {-3,-5,-8,-2,-11,-12,-13,-1};FindMaxAndSecMax(arr2,0,7,nMax,nMin);cout<<"Max: "<<nMax<<endl;//-1cout<<"Min: "<<nMin<<endl;//-2system("pause");return 1;}