給定兩個字符序列X{x1,x2,x3,...,xm}和Y{y1,y2,...,yn}
設最長公共子序列爲Z{z1,z2...zk},那麼,以下結論成立:
C(i, j)的遞推方程如下:
C(x, y) = 0 ,i=0,j=0
= C(i-1 ,j-1)+1i,j>0 , xi = yj
= max{ C(i-1 , j) , C(i , j-1)}i,j>0 , xi!= yj
網上的一張圖比較清晰地反映尋找過程,其實基本很多算法書都是參考這樣的圖~~~
public static int Length(char [] strX,char[]strY){ final int MAX = 1000; int [][] C = new int[MAX][MAX]; int [][] B = new int[MAX][MAX]; int m = strX.length+1; int n = strY.length+1; for(int i = 0 ; i < m ; i++){ C[i][0] = 0;//初始化第一行 } for(int j = 0 ; j < n ; j++){ C[0][j] = 1;//初始化第一列 } for(int i = 1 ; i < m ; i++){ for(int j = 1 ; j < n ; j++){ //i-1是因爲字符數組從第0個字符開始比較 if(strX[i-1]==strY[j-1]){ C[i][j] = C[i-1][j-1]+1; B[i][j] = 1;//C[i][j]從C[i-1][j-1]得到 }else if(C[i-1][j] >= C[i][j-1]){ C[i][j] = C[i-1][j]; B[i][j] = 2;//從C[i-1][j]得到 }else{ C[i][j] = C[i][j-1]; B[i][j] = 3;//從C[i][j-1]得到 } } } return C[m-1][n-1]; } //在實際的算法運行中,可以將B數組刪去,可以用C[i-1][j-1]保存B[i][j]的值 public static void PrintCS(int[][]B,char []X, int i, int j){ if(i==0 || j==0) return ; if(B[i][j] == 1){ PrintCS(B,X,i-1,j-1); System.out.print(X[i-1]); }else if(B[i][j] == 2){ PrintCS(B,X,i-1,j); }else{ PrintCS(B,X,i,j-1); } } PrintCS(B,strX,m-1,n-1);//注意m-1和n-1,因爲m和n都等於length+1
B[i][j]=1,C[i][j]從C[i-1][j-1]得到;B[i][j]=2,從C[i-1][j]得到;B[i][j]=3,從C[i][j-1]得到