LCS problem (longest common subsequence)-Dynamic planning implementation

Problem Description:

Problem "The longest common character sequence for two-character sequences

Attention:

Characters that do not require a substring (string one) must appear consecutively in string two.

Thinking Analysis: The optimal substructure and overlapping sub-problem properties have, so to take the dynamic programming of the longest common sub-sequence structure

Set Sequence x=

Recursive structure of xm-1= sub-problem

The optimal substructure property of the longest common sub-sequence problem is known, to find out the x=

By this recursive structure it is easy to see that the longest common subsequence problem has sub-problem overlapping properties.

For example, when calculating the longest common subsequence of x and Y, you might want to calculate the longest common subsequence of X and Yn-1 and Xm-1 and Y. These two sub-problems contain a common sub-problem, the longest common subsequence that computes Xm-1 and Yn-1.

与矩阵连乘积最优计算次序问题类似，我们来建立子问题的最优值的递归关系。用c[i,j]记录序列Xi和Yj的最长公共子序列的长度。其中Xi=<x1, x2, …, xi>，Yj=<y1, y2, …, yj>。

When i=0 or j=0, the empty sequence is the longest common sub-sequence of Xi and YJ, so c[i,j]=0. In other cases, the recursive relationship can be established by the theorem as follows:

Code:

 Public classCOMSUBSTR { Public Static void Main(string[] Arg) {String A ="Blog.csdn.net"; String B ="Csdn.blogt";      Comsubstring (A, b); }Private Static void comsubstring(String str1, String str2) {Char[] A = Str1.tochararray ();Char[] B = Str2.tochararray ();intA_length = A.length;intB_length = B.length;int[] LCS =New int[A_length +1][b_length +1];//Initialize array         for(inti =0; I <= b_length; i++) { for(intj =0; J <= A_length; J + +) {Lcs[j][i] =0; }          } for(inti =1; I <= a_length; i++) { for(intj =1; J <= B_length; J + +) {if(A[i-1] = = B[j-1]) {Lcs[i][j] = lcs[i-1][j-1] +1; }if(A[i-1]! = b[j-1]) {Lcs[i][j] = lcs[i][j-1] > Lcs[i-1][J]? Lcs[i][j-1]: Lcs[i-1][J]; }              }          }//output array results to observe         for(inti =0; I <= a_length; i++) { for(intj =0; J <= B_length; J + +) {System. out. Print (lcs[i][j]+","); } System. out. println (""); }Minimum common string constructed by array        intMax_length = Lcs[a_length][b_length];Char[] Comstr =New Char[Max_length];intI =a_length, J =b_length; while(max_length>0){if(lcs[i][j]!=lcs[i-1][j-1]){if(lcs[i-1][j]==lcs[i][j-1]){//Two characters equal, for common characterscomstr[max_length-1]=a[i-1];                      max_length--;                  i--;j--; }Else{//Take the older elders as the longest common sub-sequence of A and B                    if(lcs[i-1][j]>lcs[i][j-1]) {i--; }Else{j--; }                  }              }Else{i--;j--; }} System. out. Print ("The longest common string is:"); System. out. print (COMSTR); }  }

Output Result:

0,0,0,0,0,0,1,2,2,2,2,0,0,0,0,0,0,1,2,3,3,3,0,0,0,0,0,0,1,2,3,4,4,0,0,0,0,0,1,1,2,3,4,4,0,1,1,1,1,1,1,2,3,4,4,0,1,2,2,2,2,2,2,3,4,4,0,1,2,3,3,3,3,3,3,4,4,0,1,2,3,4,4,4,4,4,4,4,0,1,2,3,4,5,5,5,5,5,5,0,1,2,3,4,5,5,5,5,5,5,0,1,2,3,4,5,5,5,5,5,5,0,1,2,3,4,5,5,5,5,5,6The longest common string is: csdn. T  

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Reference:
http://blog.csdn.net/v_JULY_v/article/details/6110269
http://www.programgo.com/article/74411986718/

LCS problem (longest common subsequence)-Dynamic planning implementation