**一. 题目描述** 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. **二. 题目分析** 此题可使用二维动态规划来解决，下表给出了直观的匹配过程：  设某一格的状态为`k[i][j]`，表示`s1[i]`或`s2[j]`，与`s3[i+j]`的匹配结果。`s3`可与`s1`和`s2`相匹配时，可分为以下两种情况： 如果`s1` 的最后一个字符等于`s3` 的最后一个字符，则`k[i][j]=k[i-1][j]`；  如果`s2` 的最后一个字符等于`s3` 的最后一个字符，则`k[i][j]=k[i][j-1]`。 因此状态转移方程如下：  `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]);` **三. 示例代码** ~~~ #include <iostream> #include <string> #include <vector> using namespace std; class Solution { public: bool isInterleave(string s1, string s2, string s3) { if (s3.size() != s1.size() + s2.size()) return false; if (s3[0] != s1[0] && s3[0] != s2[0]) return false; vector<vector<bool> > k(s1.size() + 1, vector<bool>(s2.size() + 1, false)); k[0][0] = true; // 边界设置 for (size_t i = 1; i <= s1.size(); ++i) k[i][0] = (s1[i - 1] == s3[i - 1]) && k[i - 1][0]; for (size_t j = 1; j <= s2.size(); ++j) k[0][j] = (s2[j - 1] == s3[j - 1]) && k[0][j - 1]; for (size_t i = 1; i <= s1.size(); ++i) { for (size_t j = 1; j <= s2.size(); ++j) { k[i][j] = ((s1[i - 1] == s3[i + j - 1]) && k[i - 1][j]) || ((s2[j - 1] == s3[i + j - 1]) && k[i][j - 1]); } } return k[s1.size()][s2.size()]; } }; ~~~   **四. 小结** 编程时要注意边界条件的问题和数组的下标问题。