Interlaving String

最后更新于:2022-04-01 22:57:09

**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: ![](https://docs.gechiui.com/gc-content/uploads/sites/kancloud/2016-01-05_568bb5f07f44b.jpg) 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 #include #include 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 > k(s1.size() + 1, vector(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()]; } }; ~~~ ![](https://docs.gechiui.com/gc-content/uploads/sites/kancloud/2016-01-05_568bb5f08c5ed.jpg) ![](https://docs.gechiui.com/gc-content/uploads/sites/kancloud/2016-01-05_568bb5f0a43fe.jpg) **四. 小结** 编程时要注意边界条件的问题和数组的下标问题。
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