The sword refers to the number that appears only once in the array of the offer series source code, and the sword refers to the offer

The sword refers to the number that appears only once in the array of the offer series source code, and the sword refers to the offer

Question 1351: The number time limit that appears only once in the array: 1 second memory limit: 32 MB special question: No submitted: 2582 solution: 758 question description: in an integer array, all numbers except two appear twice. Write a program to find the numbers that appear only once. Input: each test case contains two rows: the first row contains an integer n, indicating the array size. 2 <= n <= 10 ^ 6. The second row contains n integers, indicating the array elements, all of which are int. Output: For each test case, only two numbers appear in the output array. The order of output numbers from small to large. Sample input: 82 4 3 6 3 2 5 5 sample output: 4 6

#include <iostream>#include<stdio.h>using namespace std;unsigned int findFirstBitIs1(int num){    int indexBit = 0;    while(((num&0x01)==0)&&(indexBit<8*sizeof(int))){        num=num>>1;        indexBit++;    }    return indexBit;}bool isBit1(int num, int indexBit){    num = num>>indexBit;    return (num&0x01);}void findNumsAppearOnce(int data[],int length,int* num1,int* num2){    if(data==NULL||length<2)        return;      int resultExclusiveOR = 0;      for(int i=0;i<length;i++){        resultExclusiveOR^=data[i];      }   unsigned int pos = findFirstBitIs1(resultExclusiveOR);    *num1=*num2=0;    for(int j=0;j<length;j++){        if(isBit1(data[j],pos)){            *num1^=data[j];        }else{            *num2^=data[j];        }    }}int main(){    int n;    while(scanf("%d",&n)!=EOF){        int* data = new int[n];        int num1=0,num2=0;        for(int i=0;i<n;i++){            scanf("%d",&data[i]);        }        findNumsAppearOnce(data,n,&num1,&num2);        int min,max;        min = num1<num2?num1:num2;        max = num1>num2?num1:num2;        printf("%d %d\n",min,max);    }    return 0;}