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.

最基础的做法： 硬比较，过不了大测试。

public class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(s3.length() != s1.length() + s2.length()) return false;
        return check(s1, 0, s2, 0, s3, 0);
    }
    public boolean check(String s1, int p1, String s2, int p2, String s3, int p3){
        if(p3 == s3.length()) return true;
        if(p1 < s1.length() && s3.charAt(p3) == s1.charAt(p1)){ 
            if(p2 < s2.length() && s3.charAt(p3) == s2.charAt(p2))
                return check(s1, p1, s2, p2 + 1, s3, p3 + 1) || check(s1, p1 + 1, s2, p2, s3, p3 + 1);
            else
                return check(s1, p1 + 1, s2, p2, s3, p3 + 1);
        }else if(p2 < s2.length() && s3.charAt(p3) == s2.charAt(p2))
            return check(s1, p1, s2, p2 + 1, s3, p3 + 1);
        else return false;
    }
}

第二遍： 采用二维DP来做。对于每个数 有一个关系：

isInterleaving(s1,len1,s2,len2,s3,len3)=(s3.lastChar == s1.lastChar)&& isInterleaving(s1,len1 -1,s2,len2,s3,len3 -1)||(s3.lastChar == s2.lastChar)&& isInterleaving(s1,len1,s2,len2 -1,s3,len3 -1)

public class Solution {
    boolean[][] map = null;
    public boolean isInterleave(String s1, String s2, String s3) {
        // Start typing your Java solution below
        // DO NOT write main() function
        // char l3 = s3.length();
        int l2 = s2.length();
        int l1 = s1.length();
        if(s3.length() != l2 + l1) return false;
        map = new boolean[l1 + 1][l2 + 1];
        map[0][0] = true;
        for(int i = 0; i < l1; i ++){
            if(s1.charAt(i) == s3.charAt(i)) map[i + 1][0] = true;
            else break;
        }
        for(int j = 0; j < l2; j ++){
            if(s2.charAt(j) == s3.charAt(j)) map[0][j + 1] = true;
            else break;
        }
        for(int i = 0; i < l1; i ++)
            for(int j = 0; j < l2; j ++){
                char c1 = s1.charAt(i);
                char c2 = s2.charAt(j);
                char c3 = s3.charAt(i + j + 1);
                if(c1 == c3) map[i + 1][j + 1] = map[i][j + 1] || map[i + 1][j + 1];
                if(c2 == c3) map[i + 1][j + 1] = map[i + 1][j] || map[i + 1][j + 1];
        }
        return map[l1][l2];
    }
}

 第三遍：

public class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
        int l1 = s1.length(), l2 = s2.length(), l3 = s3.length();
        if(s1.length() + s2.length() != s3.length()) return false;
        boolean[][] arr = new boolean[l1 + 1][l2 + 1];
        arr[0][0] = true;
        for(int i = 0; i < l1 && arr[i][0]; i ++){
            if(s1.charAt(i) == s3.charAt(i)) arr[i + 1][0] = true; 
        }
        for(int i = 0; i < l2 && arr[0][i]; i ++){
            if(s2.charAt(i) == s3.charAt(i)) arr[0][i + 1] = true; 
        }
        for(int i = 1; i <= l1; i ++){
            for(int j = 1; j <= l2; j ++){
                if(arr[i][j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1)) arr[i][j] =  true;
                else if(arr[i - 1][j] && s1.charAt(i - 1) == s3.charAt(i + j - 1)) arr[i][j] = true;
                else arr[i][j] = false;
            }
        }
        return arr[l1][l2];
    } 
}