【二维dp_右下递推】interleaving、Distinct Subsequences、字符串交错、min edit distance、longest common sequence(LCS)_longest common squence edit distance-优快云博客

本文链接：https://blog.youkuaiyun.com/brandohero/article/details/38343077

本文深入探讨了字符串交错和编辑距离的概念，通过具体示例解析如何判断两个字符串是否可以由另一个字符串交错组成，以及如何计算两个字符串之间的最小编辑距离。文章详细介绍了算法实现过程及其实现代码。

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1 Interleaving String

Total Accepted: 11803 Total Submissions: 62216 My Submissions

Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.

For example,
Given:
s1 = "aabcc",
s2 = "dbbca",

When s3 = "aadbbcbcac", return true.
When s3 = "aadbbbaccc", return false.

注意右下递推中边界条件：i=0? j=0?

class Solution {
public:
    bool isInterleave(string s1, string s2, string s3) {
        int n=s1.length(),m=s2.length();
        if(n+m!=s3.length()) return false;///@@@error: s3.length can be not qual to s1's + s2's
        vector<vector<bool> > s(2,vector<bool>(m+1,false));//
        s[0][0]=true;
        for(int i=0;i<=n;i++){//  s3[i+j-1]
            for(int j=0;j<=m;j++){
                if(i==0&&j==0) continue;
                s[i%2][j]=(i>0 && s[(i+1)%2][j] && s1[i-1]==s3[i+j-1] ) || (j>0 && s[i%2][j-1] && s2[j-1]==s3[i+j-1]);//右下递推
            }      //@@@error: s[(i-1)%2] ,when i-1=-1,-1%2==-1,this cause array index overflow
        }
        return s[n%2][m];
    }
};

2 Distinct Subsequences

Total Accepted: 12288 Total Submissions: 49861 My Submissions

Given a string S and a string T, count the number of distinct subsequences of T in S.

A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not).

Here is an example:
S = "rabbbit", T = "rabbit"

Return 3.

class Solution {
public:
    int numDistinct(string S, string T) {
        int n=S.length(),m=T.length();
        if(n==0||m==0) return 0;
        vector<vector<int> > cnt(2,vector<int>(n,0));
        for(int i=0;i<m;i++){
            cnt[i%2]=vector<int>(n,0);
            for(int j=0;j<n;j++){
                cnt[i%2][j]=j>0?cnt[i%2][j-1]:0;//@error: j>=1, not j>1
                if(T[i]==S[j]){///@@error: T[i]==S[j] fault as S[i]==T[j]. remember here T is the pattern, S is the matched str
                    int t=(i>0&&j>0)?cnt[(i+1)%2][j-1]:i==0?1:0;//@error: i==0 fault as i==0&&j==0. i==0 so the previous T[0~i-1] matches S[0~j-1]
                    cnt[i%2][j]+=t;//@@@error:+=cnt[(i+1)%2][j-1];
                }
            }
        }
        return cnt[(m-1)%2][n-1];
    }
};