97 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.

和edit distance一样给出dp和memorized search两种方法。memorized search速度更快。

public boolean isInterleave(String s1, String s2, String s3) {

    if ((s1.length()+s2.length())!=s3.length()) return false;

    boolean[][] matrix = new boolean[s2.length()+1][s1.length()+1];

    matrix[0][0] = true;

    for (int i = 1; i < matrix[0].length; i++){
        matrix[0][i] = matrix[0][i-1]&&(s1.charAt(i-1)==s3.charAt(i-1));
    }

    for (int i = 1; i < matrix.length; i++){
        matrix[i][0] = matrix[i-1][0]&&(s2.charAt(i-1)==s3.charAt(i-1));
    }

    for (int i = 1; i < matrix.length; i++){
        for (int j = 1; j < matrix[0].length; j++){
            matrix[i][j] = (matrix[i-1][j]&&(s2.charAt(i-1)==s3.charAt(i+j-1)))
                    || (matrix[i][j-1]&&(s1.charAt(j-1)==s3.charAt(i+j-1)));
        }
    }

    return matrix[s2.length()][s1.length()];
}

public class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
        char[] c1 = s1.toCharArray(), c2 = s2.toCharArray(), c3 = s3.toCharArray();
        int m = s1.length(), n = s2.length();
        if(m+n != c3.length) return false;
        return dfs(c1,c2,c3,0,0,0, new boolean[m+1][n+1]);
        
    }
    // 使用invalid可以避免boolean初始化的false和结果false的混淆。
    public boolean dfs(char[] c1, char[] c2, char[] c3, int i, int j, int k, boolean[][] invalid){
        if(invalid[i][j]) return false;
        if(k == c3.length) return true;
        boolean valid = 
            i < c1.length && c1[i] == c3[k] && dfs(c1,c2,c3,i+1,j,k+1, invalid) ||
            j < c2.length && c2[j] == c3[k] && dfs(c1,c2,c3,i,j+1,k+1, invalid);
        if(!valid) invalid[i][j] = true;
        return valid;
    }
}