LeetCode:Interleaving String

標籤：style   class   blog   code   strong   2014   

     Given s1, s2,s3, find whether s3 is formed by the interleaving of s1 ands2.

For example,

Given:

s1 = "aabcc",

s2 = "dbbca",

When s3 = "aadbbcbcac", return true.

When s3 = "aadbbbaccc", return false.

解題思路:

    如果熟悉動態規劃的LCS和ED問題的話，不難看出這是個dp題目.首先，我們定義如下狀態:

dp[i+1][j+1]:表示s1[0...i]與s2[0...j]能否交替形成s3[0...i+j+1]部分.

狀態轉移方程:

    dp[i+1][j+1] = (dp[i][j+1] && s1[i] == s3[i+j+1]) | (dp[i+1][j] && s2[j] == s3[i+j+1]);

解題代碼:

class Solution {public:    bool isInterleave(string s1, string s2, string s3)     {        int m = s1.size(), n = s2.size();        if (m + n != s3.size())            return false;        bool dp[m+1][n+1];        dp[0][0] = true;        //初始化邊界.        for (int i = 0; i < n; ++i)            dp[0][i+1] = dp[0][i] && s2[i] == s3[i];        for (int i = 0; i < m; ++i)            dp[i+1][0] = dp[i][0] && s1[i] == s3[i];        for (int i = 0; i < m; ++i)            for (int j = 0; j < n; ++j)                dp[i+1][j+1] = (dp[i][j+1] && s1[i] == s3[i+j+1]) | (dp[i+1][j] && s2[j] == s3[i+j+1]);        return dp[m][n];    }};