vice versa coding for value and its variables-CodePudding

I have 2 variables A and B if A =1 then B should B=2 and if A=2 then B should B=1 Like this, there are 3 pairs 1-2,3-4,5-6 What's the best way of making a code instead of just if-else

CodePudding user response：

It is possible to use simple addition and subtraction to get the other element of the two (x, x 1):

int a = 1; // the other is 2, sum is 3
int b = 3 - a;  // if a = 2, b = 1

int c = 3; // the other is 4, sum is 7
int d = 7 - c; // if c = 4, d = 3

int m = 5; // the other is 6, sum is 11
int n = 11 - m;

Another approach could be using the following logic:

else b = a - 1

So, an array could be used to provide /- 1:

int[] signs = {-1, 1}
int b = a   signs[a % 2];
a = 1, b = 2; a = 2, b = 1
a = 3, b = 4; a = 4, b = 3;
etc.

CodePudding user response：

One simple solution is a table lookup. In an array for each possible value of a I store the corresponding value of b:

private static final int[] B_PER_A = { -1, 2, 1, 4, 3, 6, 5 };

Since array indices always start at 0 in Java, we need to put a dummy value at index 0. This value is never used (or should never be, at least).

Let’s try it out:

    for (int a = 1; a <= 6; a  ) {
        int b = B_PER_A[a];
        System.out.format("a: %d; b: %d.%n", a, b);
    }

Output:

a: 1; b: 2.
a: 2; b: 1.
a: 3; b: 4.
a: 4; b: 3.
a: 5; b: 6.
a: 6; b: 5.

Generalized to more than 3 pairs

If you need to handle a variable number of pairs, resort to math.

public static int calcB(int a) {
    // 0-based index of pair (0 = 1-2, 1 = 3-4, etc.)
    int pairNumber = (a - 1) / 2;
    // a   b for given pair
    int pairSum = 4 * pairNumber   3;
    int b = pairSum - a;
    return b;
}

In each pair the sum is equivalent to 3 modulo 4. I am exploiting this fact in finding the sum for a given pair. When I subtract a from that sum, I get b. Let’s see that demonstrated too:

    for (int a = 1; a <= 8; a  ) {
        int b = calcB(a);
        System.out.format("a: %d; b: %d.%n", a, b);
    }

a: 1; b: 2.
a: 2; b: 1.
a: 3; b: 4.
a: 4; b: 3.
a: 5; b: 6.
a: 6; b: 5.
a: 7; b: 8.
a: 8; b: 7.

The latter solution is more complicated and harder to read. So if you always have got three pairs, no more, no less, I recommend the simpler table lookup presented first.