Code analysis of XOR in Java

There is an XOR operator in the bit operators of Java, which is represented by the symbol (^). Its operation rule is: in the bits of two operands, if they are the same, the result is 0, and if they are different, the result is 1. Here is an example:

The running result is: I ^ J = 13

By analyzing the above program, I = 15 to binary is 1111, j = 2 to binary is 0010, 1101 is obtained according to the operation rules of XOR, and 13 is converted to decimal

Using this rule, we can flexibly apply it to some algorithms. For example, suppose there are 2K + 1 numbers, of which 2K are the same. You need to find the different numbers, such as 2, 3, 4, 4, 3, 5, 6, 6 and 5. We can use the XOR operator to write as follows:

The result is that the number that appears only once is 2

We just skillfully use the rules of XOR operator to get the principle that a number and 0 are XOR or itself, and a number and its own XOR are 0.

The above calculation method: v = 2 ^ 3 ^ 4 ^ 4 ^ 3 ^ 5 ^ 6 ^ 6 ^ 5;

According to the exchange law and the above rules

You can deduce the number that only appears once (2K of which are the same if the preconditions are met)

summary

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