-----Edit by ZhuSenlin HDU

求一個數組的最長遞減子序列比如{9,4,3,2,5,4,3,2}的最長遞減子序列為{9,5,4,3,2}

分析：典型的動態規劃題目，對每一個數計算由它開始的最大遞減子序列的個數，並存放到一張對映表中。例如對陣列a[n]有

……

然後利用求得的對映表及最大子序列個數獲取原陣列中的元素。對於{9,4,3,2,5,4,3,2}我們求得最大子序列個數為nMaxLen=5，表為pTable={5,3,2,1,4,3,2,1}。那麼pTable中以此找出nMaxLen,nMaxLen-1,…,1對應的原陣列的值即為最大遞減子序列。對應的為{9,5,4,3,2}.複雜度為O(n²

程式碼如下

#include <iostream>
#include <cstring>
using namespace std;

int Fun(int aIn[],int pTable[],int nLen)
{
	int nMaxLen = 0;
	for(int i = nLen-1; i >= 0; --i) {
		int nMax = 0;
		for(int j = i+1; j < nLen; ++j) {
			if(aIn[j] < aIn[i]) {
				nMax = nMax < pTable[j] ? pTable[j] : nMax;
			}
		}
		pTable[i] = 1+nMax;
		nMaxLen = nMaxLen<pTable[i] ? pTable[i] : nMaxLen;
	}

	return nMaxLen;
}

void PrintMaxSubsequence(int aIn[], int pTable[], int nMaxLen, int nLen)
{
	for(int i = 0,j=0; i < nLen; ++i) {
		if(pTable[i] == nMaxLen){
			cout << aIn[i] << " ";
			nMaxLen--;
		}
	}
	cout << endl;
}
 

測試程式碼如下：

int main()
{
	int aIn[] = {9,4,3,2,5,4,3,2};
	int nLen = sizeof(aIn)/sizeof(int);
	int* pTable = new int[nLen];
	memset(pTable,0,nLen*sizeof(int));
	int nMaxLen = Fun(aIn,pTable,nLen);
	cout << nMaxLen << endl;
	PrintMaxSubsequence(aIn,pTable,nMaxLen,nLen);
	delete [] pTable;
	return 0;
}