动态规划-专题

LCS 最长公共子序列

class Solution {
public:
    /**
     * @param A, B: Two strings.
     * @return: The length of longest common subsequence of A and B.
     */
    int longestCommonSubsequence(string A, string B) {
        // write your code here
        int lena=A.length(),lenb=B.length();
        int dp[lena][lenb];
        memset(dp,0,sizeof(dp));
        for(int i=0;i<lena;i++)
        {
            if(A[i]==B[0]) dp[i][0]=1;
        }
        for(int j=0;j<lenb;j++)
        {
            if(B[j]==A[0]) dp[0][j]=1;
        }
        for(int i=1;i<lena;i++)
        {
            for(int j=1;j<lena;j++)
            {
                if(A[i]==B[j]) dp[i][j]=dp[i-1][j-1]+1;
                else dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
            }
        }
        return dp[lena-1][lenb-1];
    }
};

最长公共子串（连续）

int longestCommonSubsequence(string A, string B)
{
    int lena=A.length(),lenb=B.length();
    int dp[lena][lenb];
    int ret=0;
    for(int i=0;i<lena;i++)
    {
        if(A[i]==B[0])
        {

            dp[i][0]=1;
            ret=1;
        }
    }
    for(int i=0;i<lenb;i++)
    {
        if(A[0]==B[i])
        {
            dp[0][i]=1;
            ret=1;
        }
    }

    for(int i=1;i<lena;i++)
    {
        for(int j=1;j<lenb;j++)
        {
            if(A[i]==B[j])
            {
                dp[i][j]=dp[i-1][j-1]+1;
                ret=max(ret,dp[i][j]);
            }
            else dp[i][j]=0;

        }
    }
    return ret;

}

　　最短编辑距离

int minDistance(string word1, string word2) {
        // write your code here
        int lena=word1.length(),lenb=word2.length();
        int dp[lena+1][lenb+1];
        for(int i=0;i<=lena;i++)
        dp[i][0]=i;
        for(int i=0;i<=lenb;i++)
        dp[0][i]=i;
        for(int i=1;i<=lena;i++)
        {
            for(int j=1;j<=lenb;j++)
            {
                int del=dp[i-1][j]+1;
                int ins=dp[i][j-1]+1;
                int chan=dp[i-1][j-1]+(word1[i-1]==word2[j-1]?0:1);
                dp[i][j]=min(min(del,ins),chan);
            }
        }
        return dp[lena][lenb];
    }