leetcode 68: Interleaving String

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Interleaving StringAug 31 '12

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.

[backtracking] TLD

public class Solution {    public boolean isInterleave(String s1, String s2, String s3) {        //init check        if(s1==null || s2==null || s3==null) return false;        if(s1.length() + s2.length() != s3.length()) return false;                return rec(s1, s2, s3);            }        private boolean rec(String s1, String s2, String s3) {        if(s1.length()==0 && s2.length()==0) return s3.length()==0;        else if(s1.length()==0) return s2.equals(s3);        else if(s2.length()==0) return s1.equals(s3);                        if(s1.charAt(0)==s3.charAt(0) && rec(s1.substring(1), s2, s3.substring(1))) return true;        else if(s2.charAt(0) == s3.charAt(0) && rec(s1, s2.substring(1), s3.substring(1))) return true;        else return false;    }}

[DP]

public class Solution {    public boolean isInterleave(String s1, String s2, String s3) {        //args check        if(s1==null || s2==null || s3==null) return false;        if(s1.length() + s2.length() != s3.length()) return false;        if(s1.length() == 0) return s2.equals(s3);        if(s2.length() == 0) return s1.equals(s3);                        int n1 = s1.length();        int n2 = s2.length();                boolean b[][] =  new boolean[n2+1][n1+1];                for(int j=1; j<=n1; j++) {            if(s1.charAt(j-1) == s3.charAt(j-1) ) b[0][j]=true;            else break;        }                for(int i=1; i<=n2; i++) {            if(s2.charAt(i-1) == s3.charAt(i-1) ) b[i][0]=true;            else break;        }                for(int i=1; i<=n2; i++) {            for(int j=1; j<=n1; j++) {                if(b[i-1][j] && s2.charAt(i-1)==s3.charAt(i+j-1)) b[i][j] = true;                else if(b[i][j-1] && s1.charAt(j-1)==s3.charAt(i+j-1)) b[i][j] = true;            }        }                return b[n2][n1];    }}