编程艺术：：交替字符串 C++语言

题目和分析来自于https://github.com/julycoding/The-Art-Of-Programming-By-July/blob/master/ebook/zh/05.04.md

题目描述

输入三个字符串s1、s2和s3，判断第三个字符串s3是否由前两个字符串s1和s2交错而成，即不改变s1和s2中各个字符原有的相对顺序，例如当s1 = “aabcc”，s2 = “dbbca”，s3 = “aadbbcbcac”时，则输出true，但如果s3=“accabdbbca”，则输出false。

分析

按照原文的分析，考察的是动态规划的知识。

不过原文里的代码，我看着有点奇怪，感觉是错的。

我试了试，dp矩阵打印出来确实是错的。

下面是我改写的，dp矩阵的第0行，还有第0列，都是单独拿出来初始化的，dp[ii][0] = true 含义是string s1[0...ii-1] 在string s2为空的情况下，和string s3[0...ii-1]相等；

同理可得dp[0][jj]的含义。

#include

#include 

using namespace std;

bool isInterleave(string &s1, string &s2, string &s3)

{

    int len1 = s1.length();

    int len2 = s2.length();

    int len3 = s3.length();

    if (len1 + len2 != len3)

    {

        return false;

    }

    if (0 == len1)

    {

        return (s2 == s3) ? true : false;

    }

    if (0 == len2)

    {

        return (s1 == s3) ? true : false;

    }

    // allocating memory to store middle values

    // we need a bias to get started

    bool ** dp = new bool*[len1+1];

    for (int ii = 0; ii < len1+1; ++ii)

    {

       dp[ii] = new bool [len2+1];

    }

    // dp[ii][jj]== true means that

    // s1[0...ii-1] and s2[0...jj-1] forming s3[0...(ii+jj-1)] is true

    // dp[0][0] means two null string form a null string, which is true

    dp[0][0] = true;

    for (int ii = 1; ii < len1+1; ++ii)

    {

        dp[ii][0] = (s1.substr(0,ii) == s3.substr(0,ii)) ? true : false;

    }

    for (int jj = 1; jj < len2+1; ++jj)

    {

        dp[0][jj] = (s2.substr(0,jj) == s3.substr(0,jj)) ? true : false;

    }

    for (int ii = 1; ii < len1+1; ++ii)

    {

        for (int jj = 1; jj < len2+1; ++jj)

        {

            if (  (dp[ii-1][jj] && s1[ii - 1] == s3[ii+jj-1]) ||

                  (dp[ii][jj-1] && s2[jj - 1] == s3[ii+jj-1]) )

            {

                dp[ii][jj] = true;

            }else

            {

                dp[ii][jj] = false;

            }

        }

    }

//    for (int ii = 0; ii < len1+1; ++ii)

//    {

//        for (int jj = 0; jj < len2+1; ++jj)

//        {

//            //check char s3[ii+jj-1]

//            if (dp[ii][jj] ||

//                    ((ii - 1 > 0) && dp[ii-1][jj] && s1[ii - 1] == s3[ii+jj-1]) ||

//                    ((jj - 1 > 0) && dp[ii][jj-1] && s2[jj - 1] == s3[ii+jj-1]) )

//            {

//                dp[ii][jj] = true;

//            }else

//            {

//                dp[ii][jj] = false;

//            }

//        }

//    }

    cout << "dp matrix:" << endl;

    for (int ii = 0; ii < len1 + 1; ++ii)

    {

        for (int jj = 0; jj < len2 + 1; ++jj)

        {

            cout << dp[ii][jj] << " ";

        }

        cout << endl;

    }

    bool result = dp[len1][len2];

    for (int ii = 0; ii < len2+1; ++ii)

    {

       delete [] dp[ii];

    }

    delete [] dp;

    return result;

}

int main()

{

    string s1 = "aabcce", s2 = "dbbca";

    string s3 = "aadbbcbcace";

    if (isInterleave(s1, s2, s3))

    {

        cout << "yes" << endl;

    }else

    {

        cout << "no" << endl;

    }

}