One day Algorithm puzzle (5)---longest common substring

• Topic

given two strings str1 and str2, returns the longest common substring of two strings. For example: str1= "1AB2345CD", str2= "12345EF", common substring is "2345"

• Analytical

The difference between the longest common substring and the longest common subsequence is that the substrings are contiguous, and the subsequence is discontinuous.

The first step is to generate a dynamic planning table. Generates a matrix DP of size m*n. DP[I][J] means how long a common substring can be as long as the last character of a common substring must be str1[i] and str2[j]. For example, the meaning of str1= "a1234b", str2= "CD1234", dp[3][4] is how long the common substring can be if it is necessary to treat str1[3] and str2[4 as the last character of the common substring. So dp[3][4]=3. If str1[i]! = str2[j], then dp[i][j]=0. The specific solution process is as follows:

1. Matrix DP The first column is dp[0..m-1][0]. If str1[i]==str2[0],dp[i][0]=1, otherwise dp[i][0]=0.

2. Matrix DP First line ibid.

3. From left to right, there are two possible scenarios from top to bottom:

1) if str1[i]! = str2[j], then dp[i][j]=0

2) if str1[i] = = Str2[j], then dp[i][j]=dp[i-1][j-1] + 1

The code to calculate the DP matrix is as follows:

1  　　  Public int[] GETDP (Char[] str1,Char[] str2) {2         int[] DP =New int[str1.length][str2.length];3          for(inti = 0; i < str1.length; i++) {4             if(Str1[i] = = Str2[0])5Dp[i][0] = 1;6         }7          for(intj = 0; J < Str2.length; J + +) {8             if(Str1[0] = =Str2[j])9DP[0][J] = 1;Ten         } One          for(inti = 1; i < str1.length; i++) { A              for(intj = 1; J < Str2.length; J + +) { -                 if(Str1[i] = =Str2[j]) { -DP[I][J] = dp[i-1][j-1] + 1; the                 } -             } -         } -         returnDP; +}

It is very easy to get the longest common substring from the DP matrix. You just need to find the largest number in the DP matrix. The code is as follows.

1  　　  Publicstring Lcst1 (String str1, String str2) {2         if(str1 = =NULL|| STR2 = =NULL|| Str1.equals ("") | | Str2.equals ("")) {3             return"";4         }5         Char[] Chs1 =Str1.tochararray ();6         Char[] Chs2 =Str2.tochararray ();7         int[] DP =GETDP (CHS1, CHS2);8         intEnd = 0;9         intMax = 0;Ten          for(inti = 0; i < chs1.length; i++) { One              for(intj = 0; J < Chs2.length; J + +) { A                 if(Dp[i][j] >max) { -End =i; -Max =Dp[i][j]; the                 } -             } -         } -         returnStr1.substring (end-max+1, end+1); +}

The above algorithm requires a matrix of size m*n, but actually can be reduced to O (1), as shown in, each slash before the calculation to generate an integer variable Len,len represents the value of the upper left position, the initial len=0. The value of each position is calculated from the left-most position of the slash to the bottom right.

The code is implemented as follows:

1  　　  Publicstring Lcst2 (String str1, String str2) {2         if(str1 = =NULL|| STR2 = =NULL|| Str1.equals ("") | | Str2.equals ("")) {3             return"";4         }5         Char[] Chs1 =Str1.tochararray ();6         Char[] Chs2 =Str2.tochararray ();7         introw = 0;//line starting with Slash8         intcol = chs2.length-1;//column starting with Slash9         intmax = 0;//Record Maximum lengthTen         intend = 0;//when the maximum length is updated, the end position of the record substring One          while(Row <chs1.length) { A             inti =Row; -             intj =Col; -             intLen = 0; the             //Traverse from (i, j) to the right -              while(I < chs1.length && J <chs2.length) { -                 if(Chs1[i]! =Chs2[j]) { -Len = 0; +}Else { -len++; +                 } A                 //record the maximum and position of the ending character at                 if(Len >max) { -End =i; -Max =Len; -                 } -i++; -J + +; in             } -             if(Col > 0) {//the column with the slash start position moves to the left first tocol--; +}Else{//row moves down after the column is moved to the leftmost position -row++; the             } *         } $         returnStr1.substring (end-max+1, end+1);Panax Notoginseng}

• Resources

Programmer Code Interview Guide left Chengyun

One day Algorithm puzzle (5)---longest common substring