# [Leedcode 97] Interleaving String

Source: Internet
Author: User

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.

` Public classSolution { Public BooleanIsinterleave (string s1, String s2, string s3) {//Dynamic programming idea, constructs a two-dimensional array, Dp[i][j] Indicates whether the S1 of the first I bit and S2 of the first J-bit meet the requirements//The state transition equation is dp[i][j]= (Dp[i-1][j]&&s1.charat (i-1) ==s3.charat (i+j-1)) | |        (Dp[i][j-1]&&s2.charat (j-1) ==s3.charat (i+j-1)); //attention to the subscript problem, the size of DP is [len1+1][len2+1], Dp[0][j] represents 0 S1 characters and J S2 characters to meet the requirements of a string        intlen1=s1.length (); intLen2=s2.length (); intlen3=s3.length (); if(LEN1+LEN2!=LEN3)return false; Booleandp[][]=New Boolean[Len1+1] [Len2+1]; dp[0][0]=true;  for(inti=1;i<=len1;i++) {dp[i][0]=dp[i-1][0]&&s1.charat (i-1) ==s3.charat (i-1); }         for(intj=1;j<=len2;j++) {dp[0][j]=dp[0][j-1]&&s2.charat (j-1) ==s3.charat (j-1); }         for(inti=1;i<=len1;i++){             for(intj=1;j<=len2;j++) {Dp[i][j]= (Dp[i-1][j]&&s1.charat (i-1) ==s3.charat (i+j-1)) | | (Dp[i][j-1]&&s2.charat (j-1) ==s3.charat (i+j-1)); }        }        returnDp[len1][len2]; }}`

[Leedcode 97] Interleaving String

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