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66 lines
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Markdown
Executable File
66 lines
2.0 KiB
Markdown
Executable File
# [477. Total Hamming Distance](https://leetcode.com/problems/total-hamming-distance/)
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## 题目
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The [Hamming distance](https://en.wikipedia.org/wiki/Hamming_distance) between two integers is the number of positions at which the corresponding bits are different.
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Now your job is to find the total Hamming distance between all pairs of the given numbers.
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**Example**:
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Input: 4, 14, 2
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Output: 6
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Explanation: In binary representation, the 4 is 0100, 14 is 1110, and 2 is 0010 (just
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showing the four bits relevant in this case). So the answer will be:
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HammingDistance(4, 14) + HammingDistance(4, 2) + HammingDistance(14, 2) = 2 + 2 + 2 = 6.
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**Note**:
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1. Elements of the given array are in the range of `0` to `10^9`
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2. Length of the array will not exceed `10^4`.
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## 题目大意
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两个整数的[汉明距离](https://baike.baidu.com/item/%E6%B1%89%E6%98%8E%E8%B7%9D%E7%A6%BB/475174?fr=aladdin)指的是这两个数字的二进制数对应位不同的数量。计算一个数组中，任意两个数之间汉明距离的总和。
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## 解题思路
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- 计算一个数组内两两元素的海明距离总和。海明距离的定义是两个数二进制位不同的总个数。那么可以把数组中的每个元素 32 位的二进制位依次扫一遍，当扫到某一位上的时候，有 k 个元素在这个位上的值是 1，n - k 个元素在这个位上的值是 0，那么在这一位上所有两两元素的海明距离是 k*(n-k) ，当把 32 位全部都扫完以后，累加出来的海明距离就是所有两两元素的海明距离。
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## 代码
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```go
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package leetcode
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func totalHammingDistance(nums []int) int {
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total, n := 0, len(nums)
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for i := 0; i < 32; i++ {
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bitCount := 0
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for j := 0; j < n; j++ {
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bitCount += (nums[j] >> uint(i)) & 1
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}
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total += bitCount * (n - bitCount)
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}
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return total
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}
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// 暴力解法超时！
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func totalHammingDistance1(nums []int) int {
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res := 0
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for i := 0; i < len(nums); i++ {
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for j := i + 1; j < len(nums); j++ {
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res += hammingDistance(nums[i], nums[j])
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}
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}
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return res
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}
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``` |