LeetCode——single-number系列

LeetCode——single-number系列

Question 1

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

Note:

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

Solution

这个题的求解用到了按位异或的操作，因为两个相同的数异或是为0的，然后按位操作又满足交换律，所以一直异或下去就能得到单独的一个数字，这有点像模拟二进制求和。

class Solution {
public:
    int singleNumber(int A[], int n) {

        int sum = 0;
        for (int i = 0; i < n; i++) {
        	sum ^= A[i];
        }
        return sum;
    }
};

进阶版本

Question 2

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?

Solution

要是如果有三进制的操作，那么这道题就很简单了，那么我们需要模拟三进制的操作，ones表示一个1的个数，twos表示两个1的个数，ones&twos结果表示三个1的个数。

class Solution {
public:
    int singleNumber(int A[], int n) {
       int ones = 0;
       int twos = 0;
       int threes;
       for (int i = 0; i < n; i++) {
           int t = A[i];
           twos |= ones & t;      // 或表示加，与表示求两个1的个数
           ones ^= t;             // 异或操作表示统计一个1的个数
           threes = ones & twos;  // 与表示三个1的个数
           ones &= ~threes;
           twos &= ~threes;
       }
       return ones;
    }
};

还有一种用额外存储空间的解法，统计每一位1的个数。

class Solution {
public:
    int singleNumber(int A[], int n) {
        int bitSum[32] = {0};
        for (int i = 0; i < n; i++) {
            int mask = 0x1;
            for (int j = 0; j < 32; j++) {
                int bit = A[i] & mask;
                if (bit != 0)
                    bitSum[j] += 1;
                mask = mask << 1;
            }
        }
        int res = 0;
        for (int i = 31; i >= 0; i--) {
            res = res << 1;
            res += (bitSum[i] % 3);
        }
        return res;
    }
};