第七周作业——最长递增子序列

最新推荐文章于 2025-02-27 21:22:45 发布

原创最新推荐文章于 2025-02-27 21:22:45 发布 · 918 阅读

CC 4.0 BY-SA版权

本文介绍了一种求解最长公共子序列（LCS）问题的算法实现，并提供了详细的C++代码示例。该算法使用动态规划方法，通过构建两个字符串的长度矩阵来找出它们之间的最长公共子序列。

#include "stdafx.h"  
#include "string.h"  
#include <iostream>  
using namespace std;  
  

enum decreaseDir {kInit = 0, kLeft, kUp, kLeftUp};  
  
void LCS_Print(int **LCS_direction,   
    char* pStr1, char* pStr2,   
    size_t row, size_t col);  
  
int LCS(char* pStr1, char* pStr2)  
{  
    if(!pStr1 || !pStr2)  
        return 0;  
  
    size_t length1 = strlen(pStr1);  
    size_t length2 = strlen(pStr2);  
    if(!length1 || !length2)  
        return 0;  
  
    size_t i, j;  
  
    int **LCS_length;  
    LCS_length = (int**)(new int[length1]);  
    for(i = 0; i < length1; ++ i)  
        LCS_length[i] = (int*)new int[length2];  
  
    for(i = 0; i < length1; ++ i)  
        for(j = 0; j < length2; ++ j)  
            LCS_length[i][j] = 0;  
  
    int **LCS_direction;  
    LCS_direction = (int**)(new int[length1]);  
    for( i = 0; i < length1; ++ i)  
        LCS_direction[i] = (int*)new int[length2];  
  
    for(i = 0; i < length1; ++ i)  
        for(j = 0; j < length2; ++ j)  
            LCS_direction[i][j] = kInit;  
  
    for(i = 0; i < length1; ++ i)  
    {  
        for(j = 0; j < length2; ++ j)  
        {  
            //之前此处的代码有问题，现在订正如下：  
            if(i == 0 || j == 0)   
            {   
                if(pStr1[i] == pStr2[j])   
                {   
                    LCS_length[i][j] = 1;   
                    LCS_direction[i][j] = kLeftUp;   
                }   
                else   
                {   
                    if(i > 0)   
                    {   
                        LCS_length[i][j] = LCS_length[i - 1][j];   
                        LCS_direction[i][j] = kUp;   
                    }   
                    if(j > 0)   
                    {   
                        LCS_length[i][j] = LCS_length[i][j - 1];   
                        LCS_direction[i][j] = kLeft;   
                    }   
                }   
            }  
    
            else if(pStr1[i] == pStr2[j])  
            {  
                LCS_length[i][j] = LCS_length[i - 1][j - 1] + 1;  
                LCS_direction[i][j] = kLeftUp;  
            }  
         
            else if(LCS_length[i - 1][j] > LCS_length[i][j - 1])  
            {  
                LCS_length[i][j] = LCS_length[i - 1][j];  
                LCS_direction[i][j] = kUp;  
            }  
        
            else  
            {  
                LCS_length[i][j] = LCS_length[i][j - 1];  
                LCS_direction[i][j] = kLeft;  
            }  
        }  
    }  
    LCS_Print(LCS_direction, pStr1, pStr2, length1 - 1, length2 - 1); //调用下面的LCS_Pring 打印出所求子串。  
    return LCS_length[length1 - 1][length2 - 1];                      //返回长度。  
}  
  
void LCS_Print(int **LCS_direction,   
    char* pStr1, char* pStr2,   
    size_t row, size_t col)  
{  
    if(pStr1 == NULL || pStr2 == NULL)  
        return;  
  
    size_t length1 = strlen(pStr1);  
    size_t length2 = strlen(pStr2);  
  
    if(length1 == 0 || length2 == 0 || !(row < length1 && col < length2))  
        return;  
  
    if(LCS_direction[row][col] == kLeftUp)  
    {  
        if(row > 0 && col > 0)  
            LCS_Print(LCS_direction, pStr1, pStr2, row - 1, col - 1);  
  
        printf("%c", pStr1[row]);  
    }  
    else if(LCS_direction[row][col] == kLeft)  
    {  
        if(col > 0)  
            LCS_Print(LCS_direction, pStr1, pStr2, row, col - 1);  
    }  
    else if(LCS_direction[row][col] == kUp)  
    {  
        if(row > 0)  
            LCS_Print(LCS_direction, pStr1, pStr2, row - 1, col);  
    }  
}  
  
int _tmain(int argc, _TCHAR* argv[])  
{  
    char* pStr1="abcde";  
    char* pStr2="acde";  
    LCS(pStr1,pStr2);  
    printf("\n");  
    system("pause");  
    return 0;  
}

程序运行结果如下所示：