LeetCode 137. Single Number II

原题链接在这里：https://leetcode.com/problems/single-number-ii/

题目：

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

题解：

这道题的关键就是要求no extra memory space. 所以就用到了bit manipulation.

用one记录到当前计算的变量为止，二进制1出现"1次"(mod 3 之后的 1)的数位.

用two记录到当前计算的变量为止，二进制1出现"2次"(mod 3 之后的 2)的数位. update by ^= (once already appeared, num appeared again).

当one和two中的某一位同时为1时表示二进制1出现3次，此时需要清零.

即用二进制模拟三进制计算, 最终ones记录的是最终结果.

Time Complexity: O(n).

Space: O(1).

AC Java:

 1 class Solution {
 2     public int singleNumber(int[] nums) {
 3         int one = 0, two = 0;
 4         for(int num : nums){
 5             two ^= one & num;
 6             one ^= num;
 7             int notThree = ~(one & two);
 8             two &= notThree;
 9             one &= notThree;
10         }
11         
12         return one;
13     }
14 }

另一方法是维护一个32位的数组，因为int 都是32位的，外层loop是对应每一位，内层loop是算所有的数在这一位上出现次数的和。

Mod 3 后剩下的就是那个只出现了一次的数。与Single Number类似.

Note:1. Check if current bit is 1 using:

nums[j] >> i & 1

但要注意的是bit operator: >>, & 都比 == operator 运算级别低，所以要加括号.

2. 给了一个array代表2进制的数，转化成10进制的数就是用

res += bitCounter[i]<<i

Time Complexity: O(n). Space: O(1).

AC Java:

 1 class Solution {
 2     public int singleNumber(int[] nums) {
 3         int res = 0;
 4         
 5         int [] bitCount = new int[32];
 6         for(int i = 0; i<32; i++){
 7             for(int j = 0; j<nums.length; j++){
 8                 bitCount[i] = (bitCount[i] + (nums[j]>>i&1))%3;
 9             }
10         }
11         
12         for(int i = 0; i<32; i++){
13             res += bitCount[i]<<i;
14         }
15         
16         return res;
17     }
18 }