Given three strings: s1, s2, s3, determine whether s3 is formed by the interleaving of s1 and s2.
Example 1:
Input:
"aabcc"
"dbbca"
"aadbbcbcac"
Output:
true
Example 2:
Input:
""
""
"1"
Output:
false
Example 3:
Input:
"aabcc"
"dbbca"
"aadbbbaccc"
Output:
false
O(n2) time or better
思路：动态规划。
dp[i][j]代表由s1的前i个字母和s2的前j个字母是否能构成当前i+j个字母。
然后状态转移即可。（看第i+j+1个是否能被s1的第i+1个构成或被s2的第j+1个构成）
public class Solution {
public boolean isInterleave(String s1, String s2, String s3) {
if (s1.length() + s2.length() != s3.length()) {
return false;
}
boolean [][] interleaved = new boolean[s1.length() + 1][s2.length() + 1];
interleaved[0][0] = true;
for (int i = 1; i <= s1.length(); i++) {
if(s3.charAt(i - 1) == s1.charAt(i - 1) && interleaved[i - 1][0])
interleaved[i][0] = true;
}
for (int j = 1; j <= s2.length(); j++) {
if(s3.charAt(j - 1) == s2.charAt(j - 1) && interleaved[0][j - 1])
interleaved[0][j] = true;
}
for (int i = 1; i <= s1.length(); i++) {
for (int j = 1; j <= s2.length(); j++) {
if(((s3.charAt(i + j - 1) == s1.charAt(i - 1) && interleaved[i - 1][j]))
|| ((s3.charAt(i + j - 1)) == s2.charAt(j - 1) && interleaved[i][j - 1]))
interleaved[i][j] = true;
}
}
return interleaved[s1.length()][s2.length()];
}
}
原文：https://www.cnblogs.com/FLAGyuri/p/12078452.html