标签：style blog color io for sp div on log
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.
直接递归实现。
1 bool isInterleave(string s1, string s2, string s3) { 2 int n1 = (int)s1.length(); 3 int n2 = (int)s2.length(); 4 int n3 = (int)s3.length(); 5 6 if (n1 + n2 != n3) { 7 return false; 8 } 9 if (n1 != 0 || n1 != 0) { 10 if (n1 && n2) { 11 if (s1[0] == s2[0] && s1[0] == s3[0]) { 12 return isInterleave(s1.substr(1), s2, s3.substr(1)) || isInterleave(s1, s2.substr(1), s3.substr(1)); 13 } 14 else if (s1[0] == s3[0]) 15 return isInterleave(s1.substr(1), s2, s3.substr(1)); 16 else 17 return isInterleave(s1, s2.substr(1), s3.substr(1)); 18 } 19 else if (n1) { 20 return s1 == s3; 21 } 22 else { 23 return s2 == s3; 24 } 25 } 26 27 return false; 28 }
结果可想而知TLE，无法AC。考虑采用迭代重写
1 bool isInterleave(string s1, string s2, string s3) { 2 int n1 = (int)s1.length(); 3 int n2 = (int)s2.length(); 4 int n3 = (int)s3.length(); 5 if (n1 + n2 != n3) return false; 6 int i = 0; 7 int j = 0; 8 9 bool flag = false; 10 stack<int> __S1; 11 stack<int> __S2; 12 stack<int> __S3; 13 14 for (int k = 0; k < n3; ++k) { 15 if ((i < n1 && s3[k] == s1[i]) && (j < n2 && s3[k] == s2[j])) { 16 if (!flag) { 17 __S1.push(i); 18 __S2.push(j); 19 __S3.push(k); 20 21 ++i; 22 } 23 else { 24 flag = false; 25 ++j; 26 } 27 } 28 else if (i < n1 && s3[k] == s1[i]) { 29 ++i; 30 } 31 else if (j < n2 && s3[k] == s2[j]){ 32 ++j; 33 } 34 else { 35 if (!__S3.empty()) { 36 k = __S3.top(); 37 --k; 38 i = __S1.top(); 39 j = __S2.top(); 40 __S3.pop(); 41 __S2.pop(); 42 __S1.pop(); 43 flag = true; 44 } 45 else return false; 46 47 } 48 } 49 50 return i == n1 && j == n2; 51 }
结果对于大集合时间缩短明显，但算法复杂度太高O(2^(N+M)), N为s1的长度，M为s2的长度。考虑Dynamic Programming以减少冗余的计算。
2. Advanced Solution
1 bool isInterleave(string s1, string s2, string s3) { 2 int n1 = (int)s1.length(); 3 int n2 = (int)s2.length(); 4 int n3 = (int)s3.length(); 5 if (n1 + n2 != n3) return false; 6 7 if (n1 == 0 || n2 == 0) { 8 if (n1 == 0) { 9 return s2 == s3; 10 } 11 else return s1 == s3; 12 } 13 14 vector<vector<int> > M(n1 + 1, vector<int>(n2 + 1, 0)); 15 M[0][0] = 1; 16 17 for (int i = 1; i <= n1; ++i) { 18 if (s1[i-1] == s3[i-1]) { 19 M[i][0] = 1; 20 } 21 else break; 22 } 23 24 for (int j = 1; j <= n2; ++j) { 25 if (s2[j-1] == s3[j-1]) { 26 M[0][j] = 1; 27 } 28 else break; 29 } 30 31 for (int i = 1; i <= n1; ++i) { 32 for (int j = 1; j <= n2; ++j) { 33 if (M[i-1][j] && s1[i-1] == s3[i+j-1]) { 34 M[i][j] = 1; 35 } 36 else if (M[i][j-1] && s2[j-1] == s3[i+j-1]) { 37 M[i][j] = 1; 38 } 39 else M[i][j] = 0; 40 } 41 42 } 43 44 return M[n1][n2]; 45 }
该算法基于以下考虑，若s3.substr(0, i+j)是s1.substr(0, i) 与s2.substr(0, j)的 interleaving string 那么只有两种可能
标签：style blog color io for sp div on log
原文地址：http://www.cnblogs.com/liew/p/4034118.html
