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.

 

思路：做过几道动态规划了，终于也有了点感觉。一次就AC了，好开心啊~

用dp[m][n]来存储s3[0 ~ m+n-1]是否是s1[0~m-1]与s2[0~n-1]交替组成的。注意，没有必要存s3的维度，因为s1[0~m-1]与s2[0~n-1]只能匹配s3的m+n个字符，即第三个维度是确定的。

dp[i][j] 只有在一下两种情况下为真

① dp[i-1][j] 为真，并且 s1[i-1] == s3[i+j-1]

② dp[i][j-1] 为真，并且 s2[j-1] == s3[i+j-1]

class Solution {public:    bool isInterleave(string s1, string s2, string s3) {        int len1 = s1.length();        int len2 = s2.length();        int len3 = s3.length();        if(len3 != len1 + len2)            return false;        vector
     
      
       > dp(len1 + 1, vector
       
        (len2 + 1, false));        dp[0][0] = true;        for(int i = 1; i < len1 + 1; i++)        {            dp[i][0] =  dp[i-1][0] && (s1[i-1] == s3[i-1]);        }        for(int j = 1; j < len2 + 1; j++)        {            dp[0][j] = dp[0][j-1] && (s2[j-1] == s3[j-1]);        }        for(int i = 1; i < len1 + 1; i++)        {            for(int j = 1; j < len2 + 1; j++)            {                dp[i][j] = (dp[i-1][j] && (s1[i-1] == s3[i+j-1]))                        || (dp[i][j-1] && (s2[j-1] == s3[i+j-1]));            }        }        return dp[len1][len2];    }};