【Lintcode】029.Interleaving String

题目：

Given three strings: s1, s2, s3, determine whether s3 is formed by the interleaving of s1 and s2.

Example

For s1 = "aabcc", s2 = "dbbca"

• When s3 = "aadbbcbcac", return true.
• When s3 = "aadbbbaccc", return false.

题解：

Solution 1 ()

class Solution { public:bool isInterleave(string s1, string s2, string s3) {int n1 = s1.size(), n2 = s2.size(), n3 = s3.size();if (n1 + n2 != n3) {return false;}vector<vector<int>> dp(n1 + 1, vector<int>(n2 + 1, false));dp[0][0] = true;for (int i = 1; i <= n1; ++i) {dp[i][0] = dp[i - 1][0] && (s1[i - 1] == s3[i - 1]);}for (int i = 1; i <= n2; ++i) {dp[0][i] = dp[0][i - 1] && (s2[i - 1] == s3[i - 1]);}for (int i = 1; i <= n1; ++i) {for (int j = 1; j <= n2; ++j) {dp[i][j] = (dp[i - 1][j] && s1[i - 1] == s3[i - 1 + j]) || (dp[i][j - 1] && s2[j - 1] == s3[j - 1 + i]);}}return dp[n1][n2];} };

Solution  2 ()

class Solution { public:bool isInterleave(string s1, string s2, string s3) {if(s1.length() + s2.length() != s3.length())return false;bool dp[s2.length() + 1];for(int i = 0; i <= s1.length(); i++){for(int j = 0; j <= s2.length(); j++){if(i == 0 && j == 0)dp[j] = true;else if(i == 0)dp[j] = (dp[j - 1] && s3[i + j - 1] == s2[j - 1]);else if(j == 0)dp[j] = (dp[j] && s3[i + j - 1] == s1[i - 1]);elsedp[j] = (dp[j] && s3[i + j - 1] == s1[i - 1]) || (dp[j - 1] && s3[i + j - 1] == s2[j - 1]);}}return dp[s2.length()];} };

　　DFS

Solution 3 ()

　　BFS

Solution 4 ()

 

转载于:https://www.cnblogs.com/Atanisi/p/6884025.html

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