给出三个字符串:s1、s2、s3，判断s3是否由s1和s2交叉构成。
样例
比如 s1 = “aabcc” s2 = “dbbca”
- 当 s3 = "aadbbcbcac"，返回 true.
- 当 s3 = "aadbbbaccc"， 返回 false.
挑战
要求时间复杂度为O(n^2)或者更好
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| public class Solution {
* Determine whether s3 is formed by interleaving of s1 and s2.
* @param s1, s2, s3: As description.
* @return: true or false.
*/
public boolean isInterleave(String s1, String s2, String s3) {
if(null == s1 || null == s2 || null == s3 || s1.length() + s2.length() != s3.length())
return false;
if(s1.length() <= 0 && s2.length() <= 0 && s3.length() <= 0)
return true;
boolean[][] common = new boolean[s1.length() + 1][s2.length() + 1];
for(int i = 1;i <= s1.length();i++)
{
if(s1.charAt(i - 1) == s3.charAt(i - 1))
{
common[i][0] = true;
}
}
for(int i = 1;i <= s2.length();i++)
{
if(s2.charAt(i - 1) == s3.charAt(i - 1))
{
common[0][i] = true;
}
}
for(int i = 1;i <= s1.length();i++)
{
for(int j = 1;j <= s2.length();j++)
{
if(s1.charAt(i - 1) == s3.charAt(i + j - 1))
{
common[i][j] = common[i - 1][j];
}
if(common[i][j])
{
continue;
}
if(s2.charAt(j - 1) == s3.charAt(i + j - 1))
{
common[i][j] = common[i][j - 1];
}
}
}
return common[s1.length()][s2.length()];
}
}
|