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.

首先考虑的是用栈记录匹配的状态，代码如下，会超时

public boolean isInterleave(String s1, String s2, String s3) {
        if(s1.length()+s2.length()!=s3.length()){
            return false;
        }
        Stack<Integer> stack1 = new Stack<Integer>();
        Stack<Integer> stack2 = new Stack<Integer>();
        stack1.push(-1);
        stack2.push(-1);
        while(!stack1.empty()){
            int index1 = stack1.pop();
            int index2 = stack2.pop();
            if(index1+index2+2==s3.length()){
                return true;
            }
            if(index1+1<s1.length()&&s1.charAt(index1+1)==s3.charAt(index1+index2+2)){
                stack1.push(index1+1);
                stack2.push(index2);
            }
            if(index2+1<s2.length()&&s2.charAt(index2+1)==s3.charAt(index1+index2+2)){
                stack1.push(index1);
                stack2.push(index2+1);
            }
        }
        return false;
    }

然后想了想之前做过的题，考虑使用动态规划，设grid[i][j][k]表示s1.substring(0,i)和s2.substring(0,j)与s3.substring(0,k)是否是interleaving string，因此可得出状态方程

grid[i][j][k] = (s1.charAt(i-1) == s3.charAt(k-1) && grid[i-1][j][k-1]) || (s2.charAt(j-1) == s3.charAt(k-1) && grid[i][j-1][k-1])

代码如下：

public boolean isInterleave(String s1, String s2, String s3) {
        boolean grid[][][] = new boolean[s1.length()+1][s2.length()+1][s3.length()+1];
        if(s1.length()+s2.length()!=s3.length()){
            return false;
        }
        if(s1.equals("")){
            return s2.equals(s3);
        }else if(s2.equals("")){
            return s1.equals(s3);
        }
        grid[0][0][0] = true;
        for(int i = 1;i<=s1.length();i++){
            grid[i][0][i] = s1.charAt(i-1)==s3.charAt(i-1)&&grid[i-1][0][i-1];
        }
        for(int i = 1;i<=s2.length();i++){
            grid[0][i][i] = s2.charAt(i-1)==s3.charAt(i-1)&&grid[0][i-1][i-1];
        }
        for(int i=1;i<=s1.length();i++){
            for(int j=1;j<=s2.length();j++){
                for(int k=1;k<=s3.length();k++){
                    grid[i][j][k] = (s1.charAt(i-1)==s3.charAt(k-1)&&grid[i-1][j][k-1])||(s2.charAt(j-1)==s3.charAt(k-1)&&grid[i][j-1][k-1]);
                }
            }
        }
        return grid[s1.length()][s2.length()][s3.length()];
    }

AC  748ms

继续思考，由于k!=i+j时grid[i][j][k]的值一定是false，因此这种情况没有意义，因此可以将k用i和j表示，这样就可以降维，将grid降维二阶矩阵。代码如下

public boolean isInterleave(String s1, String s2, String s3) {
        boolean grid[][] = new boolean[s1.length()+1][s2.length()+1];
        if(s1.length()+s2.length()!=s3.length()){
            return false;
        }
        if(s1.equals("")){
            return s2.equals(s3);
        }else if(s2.equals("")){
            return s1.equals(s3);
        }
        grid[0][0] = true;
        for(int i = 1;i<=s1.length();i++){
            grid[i][0] = s1.charAt(i-1)==s3.charAt(i-1)&&grid[i-1][0];
        }
        for(int i = 1;i<=s2.length();i++){
            grid[0][i] = s2.charAt(i-1)==s3.charAt(i-1)&&grid[0][i-1];
        }
        for(int i=1;i<=s1.length();i++){
            for(int j=1;j<=s2.length();j++){
                for(int k=1;k<=s3.length();k++){
                    grid[i][j] = (s1.charAt(i-1)==s3.charAt(i+j-1)&&grid[i-1][j])||(s2.charAt(j-1)==s3.charAt(i+j-1)&&grid[i][j-1]);
                }
            }
        }
        return grid[s1.length()][s2.length()];
    }

AC  660ms