LeetCode Single Number III

最新推荐文章于 2020-05-02 20:05:14 发布

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本文介绍了一种线性时间复杂度的算法，用于从数组中找出仅出现一次的两个元素。通过位操作实现常数空间复杂度，适用于需要高效查找的数据处理场景。

Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.

For example:

Given nums = [1, 2, 1, 3, 2, 5], return [3, 5].

Note:
The order of the result is not important. So in the above example, [5, 3] is also correct.
Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

func singleNumber(nums []int) []int {
    diff := 0
    for _, num := range nums {
        diff = diff ^ num
    }
    diff = -diff & diff
    result := make([]int, 2)
    for _, num := range nums {
        if diff & num == 0 {
            result[0] = result[0] ^ num
        } else {
            result[1] = result[1] ^ num
        }
    }
    return result
}