LeetCode(97) Interleaving String解题报告

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.

解题思路:

来自http://blog.csdn.net/u011095253/article/details/9248073

但是递归解法会超时，可以参考下；
下面DP解法也不是最优的，后续会修改。

public class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
        if(s1.length() + s2.length() != s3.length())
            return false;
        return assist(s1,0,s2,0,s3,0);
    }
    public boolean assist(String s1,int t1,String s2,int t2,String s3,int t3){
        if(t3 == s3.length())
            return true;
        if(t1 == s1.length()) return s2.substring(t2).equals(s3.substring(t3));
        if(t2 == s2.length()) return s1.substring(t1).equals(s3.substring(t3));
        if(s1.charAt(t1) == s3.charAt(t3) && s2.charAt(t2) == s3.charAt(t3))
            return assist(s1,t1+1,s2,t2,s3,t3+1) || assist(s1,t1,s2,t2+1,s3,t3+1);
        if(s1.charAt(t1) == s3.charAt(t3))
            return assist(s1,t1+1,s2,t2,s3,t3+1);
        if(s2.charAt(t2) == s3.charAt(t3))
            return assist(s1,t1,s2,t2+1,s3,t3+1);
        return false;
    }
}

DP解法:

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