LeetCode Interleaving String

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.

递归超时

class Solution {public:bool isInterleave(string s1, string s2, string s3) {if (s1.size() + s2.size() != s3.size())return false;return matchOrNot(s1, 0, s2, 0, s3, 0);}bool matchOrNot(string &s1, int p1, string &s2, int p2, string &s3, int p3) {if (p3 == s3.size())return true;if (p1 == s1.size())return s2.substr(p2) == s3.substr(p3);if (p2 == s2.size())return s1.substr(p1) == s3.substr(p3);if (s1[p1] == s3[p3] && s2[p2] == s3[p3])return matchOrNot(s1, p1 + 1, s2, p2, s3, p3 + 1) || matchOrNot(s1, p1, s2, p2 + 1, s3, p3 + 1);else if (s1[p1] == s3[p3])return matchOrNot(s1, p1 + 1, s2, p2, s3, p3 + 1);else if (s2[p2] == s3[p3])return matchOrNot(s1, p1, s2, p2 + 1, s3, p3 + 1);elsereturn false;}};

DP正解：

class Solution {public:bool isInterleave(string s1, string s2, string s3) {unsigned len1 = s1.size();unsigned len2 = s2.size();if (len1 + len2 != s3.size())return false;vector<vector<bool> > array(len1 + 1, vector<bool>(len2 + 1, false));array[0][0] = true;for (int i = 1; i < len1 + 1; ++i) {if (s1[i - 1] == s3[i - 1])array[i][0] = true;elsebreak;}for (int j = 1; j < len2 + 1; ++j) {if (s2[j - 1] == s3[j - 1])array[0][j] = true;elsebreak;}for (int i = 1; i < len1 + 1; ++i) {for (int j = 1; j < len2 + 1; ++j) {if (array[i - 1][j] && s1[i - 1] == s3[i + j - 1])array[i][j] = true;else if (array[i][j - 1] && s2[j - 1] == s3[i + j - 1])array[i][j] = true;}}return array[len1][len2];}};