97. Interleaving String (String; DP)

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.

思路：将最终要求的是否能够interleave作为状态；怎么样能被interleave呢？答案是s3的每个字符，从头开始到该字符能被s1的前i个字符及s2的前j个字符表示，所以状态是个二维数组。

class Solution {
public:
bool isInterleave(string s1, string s2, string s3) {
if(s1.empty())
{
if(s2 == s3) return true;
else return false;
}
else if(s2.empty())
{
if(s1==s3) return true;
else return false;
}

int l1 = s1.length();
int l2 = s2.length();
int l3 = s3.length();
if(l3!=l1+l2) return false;

bool** hit = new bool* [l1+1]; //状态hit[i][j]表示s3[i+j-1]由s1的前i个字符和s2的前j个字符组成
for(int i = 0; i<=l1; i++)
{
hit[i] = new bool [l2+1];
}

//initialize state
hit[0][0] = true;
for ( int i = 1; i<=l1; i++)
{
if(s1[i-1] == s3[i-1])
{
hit[i][0] = true;
}
else
{
for(int j = i; j <= l1; j++){
hit[j][0] = false;
}
break;
}
}
for ( int i = 1; i<=l2; i++)
{
if(s2[i-1] == s3[i-1])
{
hit[0][i] = true;
}
else
{
for(int j = i; j <= l2; j++){
hit[0][j] = false;
}
break;
}
}

//状态转移
for ( int i = 1; i<=l1; i++)
{
for( int j = 1; j<=l2; j++)
{
if(s1[i-1]==s3[i+j-1])
{
hit[i][j] = hit[i-1][j];
}
if(s2[j-1]==s3[i+j-1])
{
hit[i][j] = hit[i][j-1] || hit[i][j];
}
}
}
return hit[l1][l2];
}
};