[LeetCode] Single Number II

Given an array of integers, every element appears three times except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

解題思路

考慮全部用二進制表示，如果我們把 第 ith 個位置上所有數字的和對3取余，那麼只會有兩個結果 0 或 1 (根據題意，3個0或3個1相加余數都為0). 因此取余的結果就是那個 “Single Number”。

public class Solution {
    public int singleNumber(int[] nums) {
        int cnt[] = new int[32];
        int single = 0;
        for (int i = 0; i < 32; i++) {
            for (int j = 0; j < nums.length; j++) {
                cnt[i] += (nums[j] >> i) & 0x1;
            }
            single |= cnt[i] % 3 << i;
        }

        return single;
    }
}

一種改進的做法是使用掩碼變量：

當第ith位出現3次時，我們就 ones 和 twos 的第ith 位設置為0。

實現代碼

Java：

// Runtime: 1 ms
public class Solution {
    public int singleNumber(int[] nums) {
        int ones = 0, twos = 0, threes = 0;
        for (int i = 0; i < nums.length; i++) {
            twos |= ones & nums[i];
            ones ^= nums[i];
            threes = ones & twos;
            //對於ones 和 twos 把出現了3次的位置設置為0 
            ones &= ~threes;
            twos &= ~threes;
        }

        return ones;
    }
}