Problem
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: 3
Example 2:
Input: [0,1,0,1,0,1,99]
Output: 99
Solution 1
/**
* @param {number[]} nums
* @return {number}
*/
var singleNumber = function(nums) {
let one = 0;
let two = 0;
let len = nums.length;
for (var i = 0; i < len; i++) {
one = (one ^ nums[i]) & ~two;
two = (two ^ nums[i]) & ~one;
}
return one;
};
Explain:
one 的二进制中每个 1 代表该位置 1 出现 1 次；
two 的二进制中每个 1 代表该位置 1 出现 2 次；
如果某位置 1 出现 3 次，则清零。
Complexity:
- Time complexity : O(n).
- Space complexity : O(1).