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?
题目的大意是:给定一个数组,数组中的数除了两个互不相同的数外,其余数两两成对出现,现在要找出这两个数
----------------------------------------------------------------------------------------------------------------------------------------------------------------------------
这道题拿在手上没什么简单的思路,参考了别人的结果“http://www.jianshu.com/p/402622b8ad43”
简要说下解题的思路,首先对所有数进行异或运算,那么相同的就会抵消掉,得到的结果就是两个相异的数a和b的异或,这个结果当作diff
然后求 (-diff & diff) 这个式子能得到二进制中只有一个1的数,这个1的位置代表着在这个位置上,a和b是不同的,那么就可以按照这个不同,
将所有数分为两组,第一组是该位置上与a相同,第二组是该位置上与b相同。
func singleNumber(nums []int) []int { diff := 0 for _, v := range nums { diff = diff ^ v } //得到的diff表示两个不同的数的按位异或的结果 diff = -diff & diff result := make([]int, 2) for _, v := range nums { if diff & v == 0 { result[0] = result[0] ^ v } else { result[1] = result[1] ^ v } } return result }
有疑问加站长微信联系(非本文作者)