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interleavingstring.java
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executable file
·69 lines (62 loc) · 2.17 KB
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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 ok[][] = new boolean[s1.length()+1][s2.length()+1];
ok[0][0]=true;
for(int i=1;i<s1.length()+1;i++){
ok[i][0] = ok[i-1][0] && s1.charAt(i-1)==s3.charAt(i-1);
}
for(int i=1;i<s2.length()+1;i++){
ok[0][i] = ok[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++){
ok[i][j] = ok[i-1][j] && s1.charAt(i-1) == s3.charAt(i+j-1);
ok[i][j] |= ok[i][j-1] && s2.charAt(j-1) == s3.charAt(j+i-1);
}
}
return ok[s1.length()][s2.length()];
}
}
/**
* A better solution with O(min(m,n)) space complexity
*
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;
}
if(s1.length()<s2.length()){
String tmp = s1;
s1 = s2;
s2 = tmp;
}
boolean ok[] = new boolean[s2.length()+1];
boolean next[] = new boolean[s2.length()+1];
ok[0]=true;
// for(int i=1;i<s1.length()+1;i++){
// ok[i][0] = ok[i-1][0] && s1.charAt(i-1)==s3.charAt(i-1);
// }
for(int i=1;i<s2.length()+1;i++){
ok[i] = ok[i-1] && s2.charAt(i-1)==s3.charAt(i-1);
}
for(int i=1;i<s1.length()+1;i++){
next[0] = ok[0] && s1.charAt(i-1)==s3.charAt(i-1);
for(int j=1;j<s2.length()+1;j++){
next[j] = ok[j] && s1.charAt(i-1) == s3.charAt(i+j-1);
next[j] |= next[j-1] && s2.charAt(j-1) == s3.charAt(j+i-1);
}
boolean tmp[] = ok;
ok = next;
next = tmp;
}
return ok[s2.length()];
}
}
*/