[Leetcode] Interleaving String

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.

比较简单的二维DP，状态方程见代码。

class Solution {public:    bool isInterleave(string s1, string s2, string s3) {        // Start typing your C/C++ solution below        // DO NOT write int main() function           if(s1.length()+s2.length()!=s3.length()) return false; //剪枝        vector<vector<bool> > answer (s1.length()+1,(vector<bool>)(s2.length()+1));                if(s1.length()==0&&s2.length()==0) return (s3.length()==0);        if(s1.length()==0) return (s2==s3);        if(s2.length()==0) return (s1==s3);                answer[0][0] = true;                for(int i=1;i<=s1.length();i++)        {            answer[i][0] = (s1[i-1]==s3[i-1])&&answer[i-1][0];        }                for(int i=1;i<=s2.length();i++)        {            answer[0][i] = (s2[i-1]==s3[i-1])&&answer[0][i-1];                    }                for(int i=1;i<=s1.length();i++)        {            for(int j=1;j<=s2.length();j++)            {                if(s1[i-1]==s2[j-1]&&s2[j-1]==s3[i+j-1])                {                    answer[i][j] = answer[i-1][j]||answer[i][j-1];                }                else if(s1[i-1]==s3[i+j-1])                {                    answer[i][j] = answer[i-1][j];                }                else if(s2[j-1]==s3[i+j-1])                {                    answer[i][j] = answer[i][j-1];                }                else                {                    answer[i][j] = false;                }            }        }                return answer[s1.length()][s2.length()];    }};