Leetcode---97. Interleaving String

Topic Link: interleaving String

Given S1, S2, S3, find whether S3 are 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.

The requirement of this problem is to detect whether the string S3 is interleaved by S1 and S2.

This is a dynamic programming topic, using Dp[i, j] to indicate whether the S1 first and the first J characters of the S2 can be interleaved to generate S3 's first i+j characters.

As for the initial values of the DP array, when J equals 0, the value of Dp[i, 0] depends on dp[i-1, 0] is true and s1[i-1] is equal to s3[i-1], i.e.

• Dp[i][0] = dp[i-1][0] && s1[i-1] = = S3[i-1]

Similarly, when I equals 0:

• DP[0][J] = dp[0][j-1] && s2[j-1] = = S3[j-1]

As for the recursive formula, it depends on whether the characters read in the current period are equal to the corresponding characters in the S3:

• If S1[I-1] is the same as s3[i+j-1], the current position S1 can be matched, so if Dp[i-1, J] is true, Dp[i, J] is true.
• If S2[J-1] is the same as s3[i+j-1], the current position S2 can be matched, so if dp[i, J-1] is true, Dp[i, J] is true.

In summary, the recursive formula is Dp[i, j] = (s1[i-1] = = S3[i+j-1] && dp[i-1, j]) | | (S2[j-1] = = S3[i+j-1] && dp[i, j-1]).

Note that if the length of the S1 plus the length of the S2 is not equal to the length of the S3, it returns false directly.

Time complexity: O (NM)

Space complexity: O (NM)

1 class Solution2 {3  Public:4     BOOL Isinterleave(string S1, string S2, string S3)5     {6         int L1 = S1.size(), L2 = S2.size(), L3 = S3.size();7 8         if(L3 != L1 + L2)9             return false;Ten  One         Vector<Vector<BOOL> > DP(L1 + 1, Vector<BOOL>(L2 + 1, false)); A  -         DP[0][0] = true; -          for(int I = 1; I <= L1; ++ I) the             DP[I][0] = DP[I - 1][0] && S1[I - 1] == S3[I - 1]; -          for(int J = 1; J <= L2; ++ J) -             DP[0][J] = DP[0][J - 1] && S2[J - 1] == S3[J - 1]; -  +          for(int I = 1; I <= L1; ++ I) -              for(int J = 1; J <= L2; ++ J) +                 DP[I][J] = (S1[I - 1] == S3[I + J - 1] && DP[I - 1][J])  A                         || (S2[J - 1] == S3[I + J - 1] && DP[I][J - 1]); at  -         return DP[L1][L2]; -     } - };

The problem has also been given the method of BFS.

Reprint please indicate source: Leetcode---97. Interleaving String

Leetcode---97. Interleaving String