題目：

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.

題解：

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 ()

?

轉載于:https://www.cnblogs.com/Atanisi/p/6884025.html

總結

以上是生活随笔為你收集整理的【Lintcode】029.Interleaving String的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。