Single Number II

最新推荐文章于 2022-07-02 13:41:47 发布

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本文介绍了一种算法，该算法能在给定的整数数组中找到仅出现一次的整数，而其他元素均出现三次。算法实现考虑了线性时间复杂度的要求，并且尽量避免使用额外内存。

Given an array of integers, every element appears three times except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

include<stdio.h>

int singleNumber(int A[], int n) {
    int bits[32];
    int i,j,result = 0;
    for(j = 0; j < 32; j++) bits[j] = 0;
    for(i = 0; i < n; i++){
        for(j = 0; j < 32; j++){
            if((1 & (A[i] >> j)) == 1) bits[j]++;
        }
    }
    for(j = 0; j < 32; j++) {
        if(bits[j] % 3 != 0){
            result += (1 << j);
        }
    }
    return result;
}

void main(){
    int A[]={2, 2, 2, 3};
    int n = 4;
    printf("%d\n", singleNumber(A, n));
}