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?
题目大意:一个数组,有两个元素出现了一次,其余元素都出现两次,让用线性时间复杂度和常量空间复杂度找出这两个数。
解题思路:考虑位操作,对所有元素做异或操作,那么最后得到的值就是要求的两个元素异或得到的,那么找出其中为1的某一位,说明这一位两个数一个为0,一个为1,以这一位为标准,把数组的元素分为两组A、B,那么要求的两个元素肯定是一个在A组,一个在B组,而对这两组各自做异或操作,就可以得到两个数就是要求的。
public class SingleNumberIII { public static void main(String[] args) { int[] res = singleNumber(new int[]{1, 2, 1, 3, 5, 2}); System.out.println(Arrays.toString(res)); } public static int[] singleNumber(int[] nums) { int[] res = new int[2]; if (nums == null || nums.length == 0) { return res; } for (int i = 0; i < nums.length; i++) { res[0] ^= nums[i]; } int pos = 0; for (int i = 0; i < 32; i++) { int offset = 1 << i; if ((res[0] & (offset)) == offset) { pos = i; break; } } res[0] = 0; int mask = 1 << pos; for (int i = 0; i < nums.length; i++) { if ((mask & nums[i]) == mask) { res[0] ^= nums[i]; } else { res[1] ^= nums[i]; } } return res; } }
时间: 2024-10-26 19:39:44