LeetCode 477. Total Hamming Distance

The Hamming distance between two integers is the number of positions at which the corresponding bits are different.

Now your job is to find the total Hamming distance between all pairs of the given numbers.

Example:

Input: 4, 14, 2

Output: 6

Explanation: In binary representation, the 4 is 0100, 14 is 1110, and 2 is 0010 (just
showing the four bits relevant in this case). So the answer will be:
HammingDistance(4, 14) + HammingDistance(4, 2) + HammingDistance(14, 2) = 2 + 2 + 2 = 6.

Note:

1. Elements of the given array are in the range of 0 to 10^9
2. Length of the array will not exceed 10^4.

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【题目分析】

给定一个int数组，求出数组中所有数值对汉明距离中总和。

【思路】

1. 两个数的汉明距离的求法

利用位操作符对两个数先求异或，然后统计结果中1的个数。在这个题目中如何求出两两来就数值对的汉明距离，时间复杂度为O(n²)。代码如下：

public class Solution {
    public int totalHammingDistance(int[] nums) {
        if(nums.length < 2) return 0;
        int result = 0;
        for(int i = 0; i < nums.length; i++) {
            for(int j = i+1; j < nums.length; j++) {
                result += hammingDistance(nums[i], nums[j]);
            }
        }
        
        return result;
    }
    
    public int hammingDistance(int num1, int num2) {
        int num = num1 ^ num2;
        int count = 0;
        for(int i=0; i < 31; i++) {
            count += 1 & num;
            num = num >> 1;
        }
        return count;
    }
}

2. 考虑每一个bit位

上面的算法虽然很容易，但是时间复杂度较高，会导致超时，如何降低时间复杂度呢？大神们的思路是考虑所有数字的同一个bit位，统计在这个bit位上出现的1的次数count，那么这个bit位在总的汉明距离中就贡献了count*(n-count)，n是数组中元素的个数。大神的分析和代码如下：

 

public class Solution {
    public int totalHammingDistance(int[] nums) {
        int total = 0, n = nums.length;
        for (int j=0;j<32;j++) {
            int bitCount = 0;
            for (int i=0;i<n;i++) 
                bitCount += (nums[i] >> j) & 1;
            total += bitCount*(n - bitCount);
        }
        return total;
    }
}