LCS问题

Posted 2020-11-20 halo-yang

分为递归和动态规划两种求解方法。

//递归求解
private static int lcs(String one,String two){
        int n = 0;
        int m = 0;
        int length = 0;
        while(one.length()!= 0 && two.length()!= 0){
            n = one.length() ;
            m = two.length() ;
            if (one.substring(n -1).equals(two.substring(m -1))) {
                String temp = one.substring(one.length()-1);
                one = one.substring(0, one.length() - 1);
                two = two.substring(0, two.length() - 1);
                return lcs(one, two) + 1;
            }else{
                return lcs(one,two.substring(0,two.length()-1))>lcs(one.substring(0,one.length()-1),two)
                        ?lcs(one,two.substring(0,two.length()-1)):lcs(one.substring(0,one.length()-1),two);
            }
        }
        return length;
    }

-----------------------------------------------------------------------------------

 

//动态规划（遍历） 求解
private static int lcser(char[] one,char[] two){
        int a = one.length;
        int b = two.length;
        int [][] arras = new int[a+1][b+1];
        for (int i=0;i<arras[0].length;i++){
            arras[0][i] = 0;
        }
        for (int j=0;j<arras.length;j++){
            arras[j][0] = 0;
        }

        for (int n=1;n <= b ;n++){
            for (int i=1;i <= a;i++){
                if (one[i-1] == two[n-1]){
                    arras[i][n] = arras[i-1][n-1] +1;
                }else{
                    arras[i][n] = Math.max(arras[i-1][n],arras[i][n-1]);
                }
            }
        }

        for (int i= 0;i<= b;i++){
            System.out.println("/n");
            for (int j= 0; j<= a;j++){
                System.out.print(arras[j][i] + "  ");
            }
        }
        return arras[a][b];
    }