@Yano 2017-12-10T01:39:57.000000Z 字数 1177 阅读 1121

LeetCode 477 Total Hamming Distance

LeetCode

题目描述

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.

题目分析

1. 对数组中所有的数进行两两组合
2. 对于每两个数的组合，求一次 Hamming Distance

代码

public int totalHammingDistance(int[] nums) {    if (nums == null || nums.length < 2) {        return 0;    }    int[] m = new int[31];// 存储对应位数，有多少个0    for(int num : nums) {        for(int i = 0; i < 31; i++) {            if ((num & (1 << i)) == 0) {                m[i]++;            }        }    }    int result = 0;    for(int i = 0; i < 31; i++) {        result += m[i] * (nums.length - m[i]);    }    return result;}

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