LeetCode-single-number-ii

最新推荐文章于 2019-08-06 15:08:25 发布

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本文介绍了一种在整数数组中找出唯一出现一次元素的算法，该算法能在线性时间内运行且不使用额外内存。通过巧妙地运用位运算，实现了对数组中每个元素出现次数的高效计数。

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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?

public class Solution
{
     
    public int singleNumber(int []nums)
    {
        int ones = 0;
        int twos = 0;
        int threes = 0;
        for(int num : nums)
        {
            twos = twos | (num&ones);
            ones = ones ^ num;
            threes = ones & twos;
            ones = ones & (~threes);
            twos = twos & (~threes);
        }
        int res = ones | twos;
        return res;
    }
}