LeetCode 137. Single Number II

原创于 2016-04-09 00:51:52 发布 · 254 阅读

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本文介绍了一种算法，用于从整数数组中找出仅出现一次的元素，而其他元素均出现三次。该算法利用位操作实现了线性时间复杂度，并且不使用额外内存，符合LeetCode的要求。

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?

Solve it according to bits of different position, we thus need to verify the 32 bits. (suppose it is a 32-bit machine)

int singleNumber(vector<int>& nums) {
    int target = 0;
    for(int i = 0; i <= 31; ++i) {
        int count = 0;
        for(int j = 0; j < nums.size(); ++j) {
            count = count + ((nums[j] >> i) & 0x1);
        }
        target = target | ((count % 3) << i);
    }
    return target;
}

Time Complexity is 32O(n). It can pass leetCode, not sure whether this is the most optimized solution though.