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.
rewrite the code
public class Solution {
/**
* Standard dynamic programming methodology.
* @param s1 -String
* @param s2 -String
* @param s3 -String which is to be checked
* @return true when s3 is formed by the interleaving of s1 and s2, false, otherwise.
* @author Averill Zheng
* @version 2014-06-27
* @since JDK 1.7
*/
public boolean isInterleave(String s1, String s2, String s3) {
if(s1.length() == 0)
return s2.equals(s3);
if(s2.length() == 0)
return s1.equals(s3);
if(s1.length() + s2.length() != s3.length())
return false;
boolean[][] check = new boolean[s1.length() + 1][s2.length() + 1];
check[0][0] = true;
for(int i = 1; i < s1.length() + 1; ++i)
check[i][0] = check[i - 1][0] && (s1.charAt(i - 1) == s3.charAt(i - 1));
for(int i = 1; i < s2.length() + 1; ++i)
check[0][i] = check[0][i - 1] && (s2.charAt(i - 1) == s3.charAt(i - 1));
for(int i = 1; i < s1.length() + 1; ++i) {
for(int j = 1; j < s2.length() + 1; ++j) {
check[i][j] = (check[i][j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1))
||(check[i - 1][j] && s1.charAt(i - 1) == s3.charAt(i + j - 1));
}
}
return check[s1.length()][s2.length()];
}
}
method: dynamical programming...
1 public class Solution { 2 public boolean isInterleave(String s1, String s2, String s3) { 3 int len1 = s1.length(), len2= s2.length(), len3 = s3.length(); 4 if(len1 == 0 || len2 == 0) 5 return s3.equals(s1) || s3.equals(s2); 6 if(len3 != len1 + len2) 7 return false; 8 int[][] paths = new int[len1 + 1][len2 + 1]; 9 paths[0][0] = 1; 10 for(int i = 1; i < len2 + 1; ++i) 11 paths[0][i] = (s2.charAt(i - 1) == s3.charAt(i - 1)) ? 1 : 0; 12 for(int i = 1; i < len1 + 1; ++i) 13 paths[i][0] = (s1.charAt(i - 1) == s3.charAt(i - 1)) ? 1 : 0; 14 for(int i = 1; i < len1 + 1; ++i){ 15 for(int j = 1; j < len2 + 1; ++j){ 16 if(paths[i][j - 1] == 1 && s3.charAt(i + j - 1) == s2.charAt(j - 1)) 17 paths[i][j] = 1; 18 else if(paths[i - 1][j] == 1 && s3.charAt(i + j - 1) == s1.charAt(i - 1)) 19 paths[i][j] = 1; 20 else 21 paths[i][j] = 0; 22 } 23 } 24 return (paths[len1][len2] == 1)? true : false; 25 } 26 }