[LeetCode]Single Number II

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本文介绍了一种算法，该算法能在给定数组中找到仅出现一次的整数，而其他元素都出现了三次。通过逐位计算并利用模3运算来实现线性时间复杂度的解决方案。

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?

Analysis:

There is a similar like the question Single Number.

From the question, we know that every element appears three times, except one only one time.

So all the three times element if we add each bit, the count in each bit must be three times of 1.

The idea is to add all in each bit, and mod %3 . we could get the number.

Java

public int singleNumber(int[] A) {
        int result = 0;
		for(int i=0;i<32;i++){
        	int count = 0;
        	for(int j=0;j<A.length;j++){
        		count += ((A[j]>>i) &1);
        	}
        	result = result | (count % 3)<<i; 
        }
		return result;
    }

c++

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