Two digits in an array that appear only once

Title:

In an array of integers, except for two digits, the other numbers appear two times. Please write the program to find the two only occurrences of the number, requirements: time complexity of O (n), Space complexity O (1)

Test Examples:

Input:

8

{2,4,3,6,3,2,5,5}

Output:

4,6

Solution:

Use XOR or solve a problem: A number xor or oneself equals 0, XOR or other number! = 0, if it is a number, then a trip through the array xor the result is what we want, and now is 2 data, then we need to carefully consider how to separate the two numbers:

First: All the numbers are different or, get a result is not equal to 0 of the number, so this result number on bits must have at least one of 1(because that 2 number is different), we start from the right side of the result to select a bit of 1, As the nth bit (this bit is 1, then it means that the first number of this position is 1, the second number is definitely 0 at this position, or vice versa, otherwise it cannot be different or! = 0); Then according to this bit, the array elements are divided into two sub-arrays, the first subarray of each number of the nth bit 1, The nth bit of each number in the second subarray is 0; Since the standard of our grouping is whether one of the numbers is 0 or 1, the number that appears two times is definitely divided into the same array, then the two parts are again different or get two numbers;

Code Implementation

Find integer number rightmost is 1 bit int findFistBitIs1 (int number) {int index = 0;        while ((index & 1) = = 0) && index <= sizeof (int) * 8) {number >>= 1;    + + index; } return index;    Test number from the right indexbit bit is 1int isbit1onindex (int no., int indexbit) {int testnumber = 1 << indexbit; Return (number & Testnumber);}    int main () {freopen ("Input.txt", "R", stdin);    int n;    CIN >> N;    int *array = new Int[n];    for (int i = 0; i < n; ++i) {cin >> array[i];    } int testnumber = 0;    for (int i = 0; i < n; ++i) {testnumber ^= array[i];    } int indexOf1 = FINDFISTBITIS1 (Testnumber);    int ans1,ans2;    ans1 = Ans2 = 0;        for (int i = 0; i < n; ++i) {if (Isbit1onindex (ARRAY[I],INDEXOF1)) {ans1 ^= array[i];        } else {ans2 ^= array[i];    }} cout << ans1 << "<< ans2 << Endl; Delete []arraY;} 

Two digits in an array that appear only once