Leetcode 477. Total Hamming Distance 自制答案

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:

Elements of the given array are in the range of 

0 

to 

10^9

Length of the array will not exceed 

10^4

.

首先奉上超时答案一枚

public class Solution {
public int totalHammingDistance(int[] nums) {
int totalDistance = 0;
for(int i = 0;i<nums.length;i++){
for(int j = i+1;j<nums.length;j++){
totalDistance = totalDistance+Integer.bitCount(nums[i]^nums[j]);
}
}
return totalDistance;
}
}时间复杂度是O(n^2),所以超时了。

后来参考了大神们的答案，一下子就想明白了。

可以从位数方面下手，假设每一位中有n个1，那么就一定会有length-n个0。所以，所有数在这一位的异或组合就有n * (length - n)种。

而题目中整形数的最大位数为32次，所以最多只要进行32 * length 次运算即可。

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