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.
Two dimentional dynamic programming
map[i][j] = true if s1[0,i-1] and s2[0,j-1] interleaves into s3[0,i+j-1].
public class Solution {
public boolean isInterleave(String s1, String s2, String s3) {
// Start typing your Java solution below
// DO NOT write main() function
if(s3.length() != s1.length() + s2.length())
return false;
boolean[][] mat = new boolean[s1.length() +1][s2.length()+1];
for(int i = 0; i < mat.length; i++){
for(int j = 0; j < mat[i].length; j++)
mat[i][j] = false;
}
mat[0][0] = true;
for(int i = 1; i <= s1.length(); i++){
if(s1.charAt(i-1) == s3.charAt(i-1)){
mat[i][0] = true;
}
else break;
}
for(int i = 1; i <= s2.length(); i++){
if(s2.charAt(i-1) == s3.charAt(i-1)){
mat[0][i] = true;
}
else break;
}
for(int i = 1; i <= s1.length(); i++){
for(int j = 1; j <= s2.length(); j++){
char c1 = s1.charAt(i-1);
char c2 = s2.charAt(j-1);
char c3 = s3.charAt(i+j-1);
if(c1 == c3) mat[i][j] |= mat[i-1][j];
if(c2 == c3) mat[i][j] |= mat[i][j-1];
}
}
return mat[s1.length()][s2.length()];
}
}
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