Bit operation 73. Two numbers that appear only once in the array

topic

Topic link: 73. Two numbers that only appear once in the array

answer

The idea is to split the question into two simple sub-questions: find a number that appears once in an array.

We need to use a property of XOR operation: $x$ ^ $x=0$ .
Meaning of the questions, in addition to the two figures are asking$x$和$y$ , the rest of the numbers appear twice or more, then the number we get after XOR all the numbers is the desired$x$和yyXOR of$y$ .

Since $x\ne y$ , so the result must not be equal to 0, that is, there must be at least one bit that is not 0 after converting it to binary.

Find the position of a bit that is not 0, and divide the given array into two arrays: a number with 1 and a number with 0.

$x$和$y$ must be located in a different array. The same number must be in the same array.

At this point, the problem has been transformed into: a number that only appears once in the array.
XOR one of the arrays and the result is $x$或$y$ .
The corresponding other number can be compared with the previous$x$与yyThe XOR value of$y$ is XORed. That$x$ ^ $x$ ^$Y=and$ .

AC code

class Solution {
public:
    vector<int> findNumsAppearOnce(vector<int>& nums) {
        int sum = 0;
        for (auto i : nums) sum ^= i;
        int k = 0;
        while (true) {
            if (!(sum & (1 << k))) k++;
            else break;
        }
        int x = 0;
        for (auto i : nums) {
            if ((1<<k) & i) x ^= i;
        }
        return vector<int> {x, x ^ sum};
    }
};