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.

//利用dp解決，dp[i][j]的狀態(tài)表示為s1[0...i] + s2[0...j]的字符串區(qū)間是否能組成s3. //那么，動(dòng)態(tài)轉(zhuǎn)移方程為: // 1） s1[i-1] == s3[i+j-1] && dp[i-1][j] = true 那么，dp[i][j] = true; // 2） s2[j-1] == s3[i+j-1] && dp[i][j-1] = true 那么，dp[i][j] = true; class Solution { public:bool isInterleave(std::string s1, std::string s2, std::string s3) {if(s1.size() + s2.size() != s3.size()) return false;std::vector<std::vector<bool>> dp(s1.size()+1,std::vector<bool>(s2.size()+1,0));dp[0][0] = 1;for (int i = 1; i < s1.size()+1; i++){if(s1[i-1] == s3[i-1] && dp[i-1][0]) dp[i][0] = true;}for (int i = 1; i < s2.size()+1; i++){if(s2[i-1] == s3[i-1] && dp[0][i-1]) dp[0][i] = true;}for (int i = 1; i < s1.size() + 1; i++){for (int j = 1; j < s2.size() + 1; j++){if(s1[i-1] == s3[i+j-1] && dp[i-1][j]) dp[i][j] = true;if(s2[j-1] == s3[i+j-1] && dp[i][j-1]) dp[i][j] = true;}}return dp[s1.size()][s2.size()];} };

轉(zhuǎn)載于:https://www.cnblogs.com/gavanwanggw/p/6880041.html

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