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leetcode[97] Interleaving String
题目给定s1和s2两个串，判断是否能合并成s3.
如果s3是s1和s2按顺序取某些数做结合的，那么就是true。例如
s1 = “abc”，s2=“def”，那么s3=“abdefc” 先去s1前两个，再取s2，再取是第三个。所以是合法的。
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.
思路一：
利用递归。
class Solution {public: bool isInterleave(string s1, string s2, string s3) { if (s1.size() == 0) return s2 == s3; if (s2.size() == 0) return s1 == s3; if (s1.size() + s2.size() != s3.size()) return false; if (s1[0] == s3[0] && s1[0] != s2[0]) return isInterleave(s1.substr(1), s2, s3.substr(1)); else if (s2[0] == s3[0] && s1[0] != s2[0]) return isInterleave(s1, s2.substr(1), s3.substr(1)); else if (s1[0] == s3[0] && s1[0] == s2[0]) return isInterleave(s1, s2.substr(1), s3.substr(1)) || isInterleave(s1.substr(1), s2, s3.substr(1)); else return false; }};
Time Limited
然后就想用动态规划了，一开始我定义dp[i][j][k],表示s1的前i个和s2的前j个是否和s3的前k个匹配。bool型的。如下：
class Solution {public: bool isInterleave(string s1, string s2, string s3) { int len1 = s1.size(), len2 = s2.size(), len3 = s3.size(); if (len1 == 0) return s2 == s3; if (len2 == 0) return s1 == s3; if (len1 + len2 != len3) return false; bool dp[len1+1][len2+1][len3+1]; for (int i = 1; i <= len3; ++i) { if (i <= len1) dp[i][0][i] = s1[i-1] == s3[i-1]; if (i <= len2) dp[0][i][i] = s2[i-1] == s3[i-1]; } for (int i = 1; i <= len1; ++i) for (int j = 1; j <= len2; ++j) { int k = i + j; if (s1[i-1] == s3[k-1] && s1[i-1] != s2[j-1]) dp[i][j][k] = dp[i-1][j][k-1]; else if (s2[j-1] == s3[k-1] && s1[i-1] != s2[j-1]) dp[i][j][k] = dp[i][j-1][k-1]; else if (s1[i-1] == s2[j-1] && s1[i-1] == s3[k-1]) dp[i][j][k] = (dp[i][j-1][k-1] || dp[i-1][j][k-1]); else dp[i][j][k] = 0; } return dp[len1][len2][len3]; }};
后面觉得不要用三维可以，可以把dp的第三维度干掉。dp[i][j]就表示s1的前i个和s2的前j个是否和s3的前i+j个匹配
class Solution {public: bool isInterleave(string s1, string s2, string s3) { int len1 = s1.size(), len2 = s2.size(), len3 = s3.size(); if (len1 == 0) return s2 == s3; if (len2 == 0) return s1 == s3; if (len1 + len2 != len3) return false; bool dp[len1+1][len2+1]; for (int i = 1; i <= len3; ++i) { if (i <= len1) dp[i][0] = s1[i-1] == s3[i-1]; if (i <= len2) dp[0][i] = s2[i-1] == s3[i-1]; } for (int i = 1; i <= len1; ++i) for (int j = 1; j <= len2; ++j) { int k = i + j; if (s1[i-1] == s3[k-1] && s1[i-1] != s2[j-1]) dp[i][j] = dp[i-1][j]; else if (s2[j-1] == s3[k-1] && s1[i-1] != s2[j-1]) dp[i][j] = dp[i][j-1]; else if (s1[i-1] == s2[j-1] && s1[i-1] == s3[k-1]) dp[i][j] = (dp[i][j-1] || dp[i-1][j]); else dp[i][j] = 0; } return dp[len1][len2]; }};
里面直接用 bool dp[len][len]这是自C99来允许的。leetcode上也允许，我在codeblock上也行，但在vs上就不行了。java就不会有这样的问题。还有为了移植性还是不那样用吧。大不了用动态数组呗。
leetcode[97] Interleaving String
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