标签：style blog io color sp for on div 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.
分析：动态规划， f[i][j]表示s3[0:i+j-1]是s1[0:i-1]和s2[0:j-1]的interleaving string。递推公式如下：
f[i][j] = (s1[i-1] == s3[i+j-1]) && f[i-1][j] || (s2[j-1] == s3[i+j-1]) && f[i][j-1].
代码如下：
class Solution { public: bool isInterleave(string s1, string s2, string s3) { if (s3.length() != s1.length() + s2.length()) return false; vector<vector<bool>> f(s1.length() + 1,vector<bool>(s2.length() + 1, true)); for (size_t i = 1; i <= s1.length(); ++i) f[i][0] = f[i - 1][0] && s1[i - 1] == s3[i - 1]; for (size_t i = 1; i <= s2.length(); ++i) f[0][i] = f[0][i - 1] && s2[i - 1] == s3[i - 1]; for (size_t i = 1; i <= s1.length(); ++i) for (size_t j = 1; j <= s2.length(); ++j) f[i][j] = (s1[i - 1] == s3[i + j - 1] && f[i - 1][j]) || (s2[j - 1] == s3[i + j - 1] && f[i][j - 1]); return f[s1.length()][s2.length()]; } };
本题目的递推版本很容易理解，但超时。
标签：style blog io color sp for on div log
原文地址：http://www.cnblogs.com/Kai-Xing/p/4154003.html
