leetcode: Interleaving String

动态规划可解：状态转换方程：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]))

也就是从后往前推 interleaving(s1, s2, s3, len1, len2, len3) =( s1.lastchar == s3.lastchar && interleaving( s1, s2, s3, len1-1, len2, len3-1)) || ( s2.lastchar == s3.lastchar && interleaving( s1, s2, s3, len1, len2-1, len3-1))

其中len3=len1+len2，所以说len3由两个变量决定，需要用二维数组

class Solution {public:    bool isInterleave(string s1, string s2, string s3) {        if( s1.size() + s2.size() != s3.size())            return false;        bool **dp = new bool *[s1.size()+1];//这里必须+1        for( int i = 0; i <= s1.size(); ++i){            dp[i] = new bool[s2.size()+1];//这里必须+1        }        dp[0][0] = true;        for( int i = 1; i <= s1.size(); ++i){            dp[i][0] = dp[i-1][0] && ( s3[i-1] == s1[i-1]);        }        for( int j = 1; j <= s2.size(); ++j){            dp[0][j] = dp[0][j-1] && ( s3[j-1] == s2[j-1]);        }        for( int i = 1; i <= s1.size(); ++i){            for( int j = 1; j <= s2.size(); ++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[s1.size()][s2.size()];    }};