【Lintcode】029.Interleaving String

题目：

Given three strings: s1, s2, s3, determine whether s3 is formed by the interleaving of s1 and s2.

Example

For s1 = "aabcc", s2 = "dbbca"

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

题解：

Solution 1 ()

class Solution {
public:
    bool isInterleave(string s1, string s2, string s3) {
        int n1 = s1.size(), n2 = s2.size(), n3 = s3.size();
        if (n1 + n2 != n3) {
            return false;
        }
        vector<vector<int>> dp(n1 + , vector<int>(n2 + , false));
        dp[][] = true;
        for (int i = ; i <= n1; ++i) {
            dp[i][] = dp[i - ][] && (s1[i - ] == s3[i - ]);
        }
        for (int i = ; i <= n2; ++i) {
            dp[][i] = dp[][i - ] && (s2[i - ] == s3[i - ]);
        }
        for (int i = ; i <= n1; ++i) {
            for (int j = ; j <= n2; ++j) {
                dp[i][j] = (dp[i - ][j] && s1[i - ] == s3[i -  + j]) || (dp[i][j - ] && s2[j - ] == s3[j -  + i]);
            }
        }
        return dp[n1][n2];
    }
};

Solution  2 ()

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

　　DFS

Solution 3 ()

　　BFS

Solution 4 ()