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477. Total Hamming Distance 总的汉明距离

时间:2018-01-20 00:22:25      阅读:181      评论:0      收藏:0      [点我收藏+]

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    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

  1. class Solution1(object):
  2. def totalHammingDistance(self, nums):
  3. """
  4. :type nums: List[int]
  5. :rtype: int
  6. """
  7. res = 0
  8. bins = []
  9. for num in nums:
  10. bins.append(‘{0:b}‘.format(num).zfill(32))
  11. for i in range(32):
  12. bitCount = 0
  13. # zeroCount
  14. for num in bins:
  15. if num[i] is ‘1‘:
  16. bitCount += 1
  17. # zeroCount = (len(nums) - bitCount)
  18. res += bitCount * (len(nums) - bitCount)
  19. return res
  20. class Solution2(object):
  21. def totalHammingDistance(self, nums):
  22. res = 0
  23. for i in range(32):
  24. bitCount = 0
  25. for num in nums:
  26. bitCount += num >> i & 1
  27. res += bitCount * (len(nums) - bitCount)
  28. return res
  29. nums = [4, 14, 2]
  30. s = Solution1()
  31. res = s.totalHammingDistance(nums)
  32. print(res)






477. Total Hamming Distance 总的汉明距离

标签:blank   pairs   format   lis   return   nta   print   ant   code   

原文地址:https://www.cnblogs.com/xiejunzhao/p/8319176.html

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