LintCode——交叉字符串

描述：给出三个字符串:s1、s2、s3，判断s3是否由s1和s2交叉构成。

样例：s1 = "aabcc" s2 = "dbbca"

　　- 当 s3 = "aadbbcbcac"，返回  true.

　　- 当 s3 = "aadbbbaccc"， 返回 false.

Java

 public class Solution {
     /**
      * @param s1: A string
      * @param s2: A string
      * @param s3: A string
      * @return: Determine whether s3 is formed by interleaving of s1 and s2
      */
     public boolean isInterleave(String s1, String s2, String s3){
         // write your code here
         int s1_len=s1.length();
         int s2_len=s2.length();
         int s3_len=s3.length();
         if(s1_len == 0){
             if(s2.equals(s3))
                 return true;
             else
                 return false;
         }
         if(s2_len == 0){
             if(s1.equals(s3))
                 return true;
             else
                 return false;
         }
         int visited[][] = new int[s1_len + 1][s2_len + 1];
         visited[0][0]=1;
         for(int i = 0;i < s1_len;i++){
             if(s1.charAt(i) == s3.charAt(i))
                 visited[i + 1][0] = 1;
             else
                 break;
         }
         for(int j = 0;j < s2_len;j++){
             if(s2.charAt(j) == s3.charAt(j))
                 visited[0][j + 1] = 1;
             else
                 break;
         }
         for(int i = 1;i < visited.length;i++){
             for(int j = 1;j < visited[0].length;j++){
                 int row = i - 1;
                 int col = j - 1;
                 if((visited[i][j - 1] == 1 &&
                     s2.charAt(col) == s3.charAt(row + col + 1)) ||
                    (visited[i - 1][j] == 1 &&
                     s1.charAt(row) == s3.charAt(row + col  + 1)))
                     visited[i][j]=1;
             }
         }
         if(visited[visited.length - 1][visited[0].length - 1] == 1)
             return true;
         return false;
     }
 }