给定三个字符串 s1, s2, s3, 验证 s3 是否是由 s1 和 s2 交错组成的。
示例 1:
输入: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
输出: true
示例 2:
输入: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
输出: false
解题思路:
动态规划,辅助数组boolean[][] p,p[i][j]表示s3[0~i+j-1]是否由s1[0~i]和s2[0~j]交错组成
p[i][j] = (p[i-1][j] && s1.charAt(i-1) == s3.charAt(i+j-1)) || (p[i][j-1] && s2.charAt(j-1) == s3.charAt(i+j-1));
Java代码:
class Solution {
public boolean isInterleave(String s1, String s2, String s3) {
if((s1.length() + s2.length()) != s3.length())
return false;
if(0 == s1.length())
return s2.equals(s3);
if(0 == s2.length())
return s1.equals(s3);
if(0 == s3.length())
return true;
boolean[][] p = new boolean[s1.length()+1][s2.length()+1];
p[0][0] = true;
for(int i = 1; i <= s1.length(); i++){
p[i][0] = p[i-1][0] && s1.charAt(i-1) == s3.charAt(i-1);
}
for(int j = 1; j <= s2.length(); j++){
p[0][j] = p[0][j-1] && s2.charAt(j-1) == s3.charAt(j-1);
}
for(int i = 1; i <= s1.length(); i++){
for(int j = 1; j <= s2.length(); j++){
p[i][j] = (p[i-1][j] && s1.charAt(i-1) == s3.charAt(i+j-1))
|| (p[i][j-1] && s2.charAt(j-1) == s3.charAt(i+j-1));
}
}
return p[s1.length()][s2.length()];
}
}