97. Interleaving String

dfs:

题意就是s1和s2是否可以交叉变成s3，一开始我使用dfs，但是超时了

然后剪枝，加入一个标记数组，如果dfs过的就不进行dfs，然后就ac了

class Solution {
public:
int mapp[500][500];
int mark = 0;
void pipei(string s1, string s2, string s3, int a, int b, int c){
if(mark) return ;
mapp[a][b] = 1;
if(a + 1 == s1.length() && b == s2.length() && c + 1 == s3.length() && s1[a] == s3[c]){
mark = 1;
return ;
}
else if(a == s1.length() && b + 1 == s2.length() && c + 1 == s3.length() && s2[b] == s3[c]){
mark = 1;
return ;
}
else if(a < s1.length() && b < s2.length()){
if(s1[a] == s3[c] && !mapp[a + 1][b])
pipei(s1, s2, s3, a + 1, b, c + 1);
if(s2[b] == s3[c] && !mapp[a][b + 1])
pipei(s1, s2, s3, a, b + 1, c + 1);
}
else if(a < s1.length()){
if(s1[a] == s3[c] && !mapp[a + 1][b])
pipei(s1, s2, s3, a + 1, b, c + 1);
}
else if(b < s2.length()){
if(s2[b] == s3[c] && !mapp[a][b + 1])
pipei(s1, s2, s3, a, b + 1, c + 1);
}
return ;
}
bool isInterleave(string s1, string s2, string s3) {
if(s1.length() + s2.length() != s3.length()) return false;
if(s1.length() == 0 && s2.length() == 0 && s3.length() == 0) return true;
memset(mapp, 0, sizeof(mapp));
pipei(s1, s2, s3, 0, 0, 0);
if(mark) return true;
else return false;
}
};

dp:

其实这题应该用dp，mapp (i,j)代表到s1 i 和s2 j都可以匹配，然后方程是

if(mapp[i - 1][j] && s1[i - 1] == s3[i + j - 1])

mapp[i][j] = true;

if(mapp[i][j - 1] && s2[j - 1] == s3[i + j - 1])

mapp[i][j] = true;

十分好理解

class Solution {
public:
bool isInterleave(string s1, string s2, string s3) {
if(s1.length() + s2.length() != s3.length()) return false;
if(s1.length() == 0 && s2.length() == 0 && s3.length() == 0) return true;
bool mapp[s1.length() + 1][s2.length() + 1];
memset(mapp, false, sizeof(mapp));

for(int i = 0; i <= s1.length(); ++ i)
for(int j = 0; j <= s2.length(); ++ j){
if(i == 0 && j == 0)
mapp[i][j] = true;
else if(i == 0){
if(mapp[0][j - 1] && s2[j - 1] == s3[j - 1])
mapp[0][j] = true;
}
else if(j == 0){
if(mapp[i - 1][0] && s1[i - 1] == s3[i - 1])
mapp[i][0] = true;
}
else{
if(mapp[i - 1][j] && s1[i - 1] == s3[i + j - 1])
mapp[i][j] = true;
if(mapp[i][j - 1] && s2[j - 1] == s3[i + j - 1])
mapp[i][j] = true;
}
}

return mapp[s1.length()][s2.length()];
}
};