97. Interleaving String

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.

递归方法：
bool isInterleave(char* s1, char* s2, char* s3) {
int n1 = strlen(s1);
int n2 = strlen(s2);
int n3 = strlen(s3);

if((n1+n2)!=n3){
return false;
}

if(n3 == 0 && n1 == 0 && n2 == 0){
return true;
}

if(s1[0] != s3[0] && s2[0] != s3[0]){
return false;
}

if(s1[0] != '\0' && s1[0] == s3[0] && s2[0]!=s3[0]){
return isInterleave(s1+1,s2,s3+1);
}

if(s2[0] != '\0' && s1[0] != s3[0] && s2[0]==s3[0]){
return isInterleave(s1,s2+1,s3+1);
}

if(s1[0] == s3[0] && s2[0] == s3[0]){
return isInterleave(s1,s2+1,s3+1)||
isInterleave(s1+1,s2,s3+1);
}

return true;
}

非递归方法
class Solution {
public:
bool isInterleave(string s1, string s2, string s3) {
int n1 = s1.size();
int n2 = s2.size();
int n3 = s3.size();

vector<bool> t(n2+1,false);
vector<vector<bool>> dp(n1+1,t);

if(n1+n2 != n3){
return false;
}

for(int i = 0; i <= n1; ++i){
for(int j = 0; j<= n2; ++j){
if(i == 0 && j == 0){
dp[i][j] = true;
}else if(i == 0){
dp[i][j] = dp[i][j-1] && s2[j-1] == s3[i+j-1];
}else if(j == 0){
dp[i][j] = dp[i-1][j] && s1[i-1] == s3[i+j-1];
}else{
dp[i][j] = (dp[i][j-1] && s2[j-1] == s3[i+j-1])||(dp[i-1][j] && s1[i-1] == s3[i+j-1]);
}
}
}

return dp[n1][n2];
}
};