java - 如何避免java中的重复代码？

Question

我只是编程的初学者。 uf 是一个 union-find 类，其方法 union 连接两个节点的根。 这段代码负责打开网格的一个站点，如果打开了任何一个邻居，则将该站点与其邻居合并。如果其邻居之一已满，则填充与该站点连接的所有节点。 这是实际的代码:

        if(i == 1){
            uf.union(len*len, xyTo1D(i,j));
            if(existAndOpen(i+1,j)){
                uf2.union(xyTo1D(i+1,j), xyTo1D(i,j));
                uf.union(xyTo1D(i,j), xyTo1D(i+1,j));
            }
            if(existAndOpen(i-1,j)){
                uf2.union(xyTo1D(i-1,j), xyTo1D(i,j));
                uf.union(xyTo1D(i,j), xyTo1D(i-1,j));
            }
            if(existAndOpen(i,j-1)){
                uf2.union(xyTo1D(i,j-1), xyTo1D(i,j));
                uf.union(xyTo1D(i,j), xyTo1D(i,j-1));
            }
            if(!(j == len && i == len)){
                if(existAndOpen(i,j+1)){
                    uf2.union(xyTo1D(i,j+1), xyTo1D(i,j));
                    uf.union(xyTo1D(i,j), xyTo1D(i,j+1));
                }
            }
        }
        else{
        if(existAndFull(i+1,j)){
            uf2.union(xyTo1D(i+1,j), xyTo1D(i,j));
            uf.union(xyTo1D(i,j), xyTo1D(i+1,j));
        }
        if(existAndFull(i-1,j)){
            uf2.union(xyTo1D(i-1,j), xyTo1D(i,j));
            uf.union(xyTo1D(i,j), xyTo1D(i-1,j));
        }
        if(existAndFull(i,j-1)){
            uf2.union(xyTo1D(i,j-1), xyTo1D(i,j));
            uf.union(xyTo1D(i,j), xyTo1D(i,j-1));
        }
        if(!(j== len && i == len)){
            if(existAndFull(i,j+1)){
                uf2.union(xyTo1D(i,j+1), xyTo1D(i,j));
                uf.union(xyTo1D(i,j), xyTo1D(i,j+1));
            }
        }
        if(existAndOpen(i+1,j)){
            uf.union(xyTo1D(i,j), xyTo1D(i+1,j));
        }
        if(existAndOpen(i-1,j)){
            uf.union(xyTo1D(i,j), xyTo1D(i-1,j));
        }
        if(existAndOpen(i,j-1)){
            uf.union(xyTo1D(i,j), xyTo1D(i,j-1));
        }
        if(!(j== len && i == len)){
            if(existAndOpen(i,j+1)){
                uf.union(xyTo1D(i,j), xyTo1D(i,j+1));
            }
        }
    }
    }

如何简化代码？

Answer 1

试试这个

boolean f1(int a, int b) { }
boolean f2(int a, int b) { }
void A(int a, int b) { }

void testAndA(BiPredicate<Integer, Integer> p, int a, int b) {
    if (p.test(a, b))
        A(a, b);
}

和

    if(x == 1){
        testAndA(this::f1, x + 1, y);
        testAndA(this::f1, x, y + 1);
    } else {
        testAndA(this::f2, x + 1, y);
        testAndA(this::f2, x, y + 1);
    }