《编程之美》读书笔记17： 2.16 求数组中最长递增子序列

http://blog.csdn.net/ztj111/archive/2008/07/31/2748152.aspx

#include "stdafx.h"#include <iostream>#include <algorithm>using namespace std;//复杂度为O(N2 + N) = O(N2)int MaxLength(int arr[], int length){int *res = new int[length];for (int i = 0; i<length; i++){res[i] = 1; //初始化默认的长度for (int j = 0; j<i; j++)//找出前面最长的序列{if (arr[i] > arr[j] && (res[j] + 1 > res[i]))res[i] = res[j] + 1;}}return *max_element(res, res + length); //查找最大的值}//O(N2),若使用二分查找为O(N*logN)int MaxLength2(int arr[], int length){int *maxV = new int[length + 1];maxV[1] = arr[0]; //数组中第一个值，边界值maxV[0] = *min_element(arr, arr+length) - 1;//数组中最小值，边界值int *LIS = new int[length];//初始最长递增序列的信息for(int i = 0; i<length; i++)LIS[i] = 1;int nMaxLIS = 1;//最长递增子序列的长度for (int i = 1; i<length; i++){int j;for (j = nMaxLIS; j>=0; j--){if(arr[i] > maxV[j]){LIS[i] = j + 1;break;}} //这个for可以换成二分查找if (LIS[i] > nMaxLIS)//更新最长信息{nMaxLIS = LIS[i];maxV[LIS[i]] = arr[i];}else if (maxV[j] < arr[i] && arr[i] < maxV[j+1]){maxV[j+1] = arr[i];}}return nMaxLIS;}int main(){int arr[] = {1, -1, 2, -3, 4, -5, 6, -7};int len = sizeof(arr) / sizeof(*arr);int res = MaxLength2(arr, len);printf("The max length is %d/n", res);system("pause");}