lecode Interleaving String

这个问题，前面思考过，当时就是用搜索的方法，此处又遇到一次，发现自己理解的太浅了

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.

1.搜索的方法(超时)

 public class Solution {
     public boolean isInterleave(String s1, String s2, String s3) {
          return isInter(s1,s2,s3,0,0,0);

     }
     public boolean isInter(String s1,String s2,String s3,int r1,int r2, int r3)
     {
         if(r3==s3.length()) return true;
         boolean ans=false;

         if(r1<s1.length()&&s1.charAt(r1)==s3.charAt(r3))
         {
            ans=isInter(s1,s2,s3,r1+1,r2,r3+1);

         }
         if(ans) return true;

         if(r2<s2.length()&&s2.charAt(r2)==s3.charAt(r3))
         {
            ans=isInter(s1,s2,s3,r1,r2+1,r3+1);
            return ans;

         }

         return false;

     }
 }

dp[i][j]表示s1前i 和s2前j个是否能组成s3的前i+j+1个,  false 不能  true 能

dp[s1.len-1][s2.len-1] 就是我们的答案

dp[i][j]=dp[i-1][j]&&s1[i]==s3[i+j+1]||                dp[i][j-1]&&s1[j]==s3[i+j+1]

 public class Solution {
     public boolean isInterleave(String s1, String s2, String s3) {
         char c1[]=s1.toCharArray();

     char c2[]=s2.toCharArray();
     char c3[]=s3.toCharArray();
         int len1=s1.length();
         int len2=s2.length();
         int len3=s3.length();
         if(len1+len2!=len3) return false;
         if(len1==0) return s2.equals(s3);
         if(len2==0) return s1.equals(s3);

         boolean dp[][]=new boolean[s1.length()+1][s2.length()+1];
         dp[0][0]=true;
         for(int i=1;i<=len1;i++)
         {

             dp[i][0]=dp[i-1][0]&&c1[i-1]==c3[i-1];

         }
         for(int j=1;j<=len2;j++)
         {

             dp[0][j]=dp[0][j-1]&&c2[j-1]==c3[j-1];
         }

         for(int i=1;i<=len1;i++)
         {

             for(int j=1;j<=len2;j++)
             {

                 dp[i][j]=dp[i-1][j]&&(c1[i-1]==c3[i+j-1]);
                 if(dp[i][j]) continue;
                   dp[i][j]=dp[i][j-1]&&(c2[j-1]==c3[i+j-1]);

             }

         }

          return dp[len1][len2];

     }

 }