题目
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.
分析
题目给定三个字符串s1 , s2 , s3，要求判定字符串s3是否由s1 和s2组合而成。（每个字符串中的字母相对顺序不可变）开始看到题目，没有解决思路。参考网友的答案，发现解决题目的两种思路。方法一：递归的思路，但是该方法对于大集合数据会出现TLE方法二：动态规划的思路根据字符串1和2，建立判定二维矩阵。
AC代码
class Solution {
public:
/*方法一：递归实现,对大数据组会TLE*/
bool isInterleave1(string s1, string s2, string s3) {
int len1 = s1.length(), len2 = s2.length(), len3 = s3.length();if (len2 == 0)
return s1 == s3;
else if (len1 == 0)
return s2 == s3;
else if (len3 == 0)
return len1 + len2 == 0;
else
{
if (s1[0] == s3[0] && s2[0] != s3[0])
return isInterleave1(s1.substr(1), s2, s3.substr(1));
else if (s1[0] != s3[0] && s2[0] == s3[0])
return isInterleave1(s1, s2.substr(1), s3.substr(1));
else if (s1[0] == s3[0] && s2[0] == s3[0])
return isInterleave1(s1.substr(1), s2, s3.substr(1)) || isInterleave(s1, s2.substr(1), s3.substr(1));
else
return false;
}//else
}/*方法二：二维动态规划*/
bool isInterleave(string s1, string s2, string s3) {
int len1 = s1.length(), len2 = s2.length(), len3 = s3.length();
if (len1 + len2 != len3)
return false;
else if (len2 == 0)
return s1 == s3;
else if (len1 == 0)
return s2 == s3;
else if (len3 == 0)
return len1 + len2 == 0;
else
{
vector<vector<int>> dp(len1 + 1, vector<int>(len2 + 1, 0));
dp[0][0] = 1;
for (int i = 1; i <= len1; ++i)
{
if (s1[i - 1] == s3[i - 1])
dp[i][0] = 1;
else
break;
}//forfor (int j = 1; j <= len2; ++j)
{
if (s2[j - 1] == s3[j - 1])
dp[0][j] = 1;
else
break;
}//forfor (int i = 1; i <= len1; ++i)
{
for (int j = 1; j <= len2; ++j)
{
if (s1[i - 1] == s3[i + j - 1])
dp[i][j] = dp[i - 1][j] || dp[i][j];
if (s2[j - 1] == s3[i + j - 1])
dp[i][j] = dp[i][j - 1] || dp[i][j];
}//for
}//for
return dp[len1][len2] == 1;
}//else
}
};
