Interleaving String

题目描述：

Given s1, s2, s3, find whether s3 is formed by the interleaving ofs1 and s2.

For example,
Given:
s1 = "aabcc",
s2 = "dbbca",

When s3 = "aadbbcbcac", return true.
When s3 = "aadbbbaccc", return false.

这个题一开始用的是dfs，超时了：

public class Solution {    public boolean isInterleave(String s1, String s2, String s3) {int n1=s1.length(),n2=s2.length(),n3=s3.length();if(n1+n2!=n3){return false;}if(n3==0)return true;int i=n1-1,j=n2-1,k=n3-1;if(i>=0&&s1.charAt(i)==s3.charAt(k)&&(j==-1||s2.charAt(j)!=s3.charAt(k))){return isInterleave(s1.substring(0,i), s2, s3.substring(0,k));}if(j>=0&&s2.charAt(j)==s3.charAt(k)&&(i==-1||s1.charAt(i)!=s3.charAt(k))){return isInterleave(s1, s2.substring(0,j), s3.substring(0,k));}if(i>=0&&s1.charAt(i)==s3.charAt(k)&&j>=0&&s2.charAt(j)==s3.charAt(k)){return isInterleave(s1.substring(0,i), s2, s3.substring(0,k))||isInterleave(s1, s2.substring(0,j), s3.substring(0,k));}return false;    }}

这个题应该用动态规划算法，dp[i][j]表示用s1的前i个字符和s2的前j个字符能不能按照规则表示出s3的前i+j个字符，如此最后结果就是dp[s1.length()][s2.length()]，判断是否为真即可。

public class Solution {    boolean dp[][];  public boolean isInterleave(String s1, String s2, String s3){          int size1 = s1.length();          int size2 = s2.length();          int size3 = s3.length();          if( size1 + size2 != size3) return false;                    dp=new boolean[size1+1][size2+1];         dp[0][0] = true;          for(int i = 1; i <= size1; ++i)              if(s1.charAt(i-1) == s3.charAt(i-1)) dp[i][0] = true;              else break;          for(int j = 1; j <= size2; ++j)              if(s2.charAt(j-1) == s3.charAt(j-1)) dp[0][j] = true;              else break;            int k;          for(int i = 1; i <= size1; ++i)              for(int j = 1; j <= size2; ++j)              {                  k = i + j;                  if(s1.charAt(i-1) == s3.charAt(k-1)) dp[i][j] = dp[i-1][j] || dp[i][j];                  if(s2.charAt(j-1) == s3.charAt(k-1)) dp[i][j] = dp[i][j-1] || dp[i][j];              }          return dp[size1][size2];      }  }