第48题 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.

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 Dynamic Programming String

Solution in Java：（Dynamic）

public class Solution {    public boolean isInterleave(String s1, String s2, String s3) {        if(s1=="")  return  s2.equals(s3);        if(s2=="") return s1.equals(s3);                int len1=s1.length(), len2 = s2.length(), len3 = s3.length();        if(len3!=len1+len2) return false;                boolean[][] opj = new boolean[len1+1][len2+1];        //opj[i][j] denotes if s1[0...i-1] and s2[0...j-1] can form s3[0...i+j-1].        //so opj[len1][len2] denotes  if s1[0...len1-1] and s2[0...len2-1] can from s3[0...len1+len2-1],        //which is the answer we need.                opj[0][0]=true;                for(int j=1; j<len2+1; j++) opj[0][j] = s2.charAt(j-1)==s3.charAt(j-1);                for(int i=1; i<len1+1; i++) opj[i][0] = s1.charAt(i-1)==s3.charAt(i-1);                for(int i=1; i<len1+1; i++){            for(int j=1; j<len2+1; j++){                opj[i][j] = opj[i-1][j]&&s1.charAt(i-1)==s3.charAt(i+j-1)|| //add char in s1                            opj[i][j-1]&&s2.charAt(j-1)==s3.charAt(i+j-1);  //add char in s2            }        }        return opj[len1][len2];    }}

Note： 只关心s3[0...i+j-1]是否能由s1[0...i-1]和s2[0...j-1]形成，不关心之前怎样形成的具体过程。

而在s3[0...i+j-2]合法的前提下，要s3[0...i+j-1]也合法，需要由s1或s2中添加一个合法字母。