Given a non-empty array of integers, every element appears three times except for one, which appears exactly once. Find that single one.Note:Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?Example 1:Input: [2,2,3,2]Output: 3Example 2:Input: [0,1,0,1,0,1,99]Output: 99难度: medium题目：给定一个非空整数数组，其中一个元素出现一次，其它元素出现三次。找出有且仅出现一次的那个元素。注意：你的算法时间复杂度为O(n)。能否实现它不使用额外的内存？思路：按位统计各个元素中位出现的次数，因为除1次出现的元素之外，其它元素都出现3次，因此按位统计位1和位0的个数应为3的倍数。最后只出现1次的那个元素的所有位都会剩下来。Runtime: 3 ms, faster than 60.00% of Java online submissions for Single Number II.class Solution { public int singleNumber(int[] nums) { int result = 0, bit = 0; for (int i = 0; i < 32; i++, bit = 0) { for (int j = 0; j < nums.length; j++) { bit += (nums[j] >> i) & 1; bit %= 3; } result |= (bit << i); } return result; }}