给定一个数组 nums ，如果 i < j 且 nums[i] > 2*nums[j] 我们就将 (i, j) 称作一个重要翻转对。
你需要返回给定数组中的重要翻转对的数量。
示例 1:
输入: [1,3,2,3,1]
输出: 2
示例 2:
输入: [2,4,3,5,1]
输出: 3
注意:
给定数组的长度不会超过50000。
输入数组中的所有数字都在32位整数的表示范围内。
链接：https://leetcode-cn.com/problems/reverse-pairs
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| class Solution(object):
def reversePairs(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
dic = {}
cnt = 0
s = []
for i in nums[::-1]:
if not s:
s.append(i*2)
else:
l, r = 0, len(s)-1
while l <= r:
mid = (l + r) >> 1
if s[mid] >= i:
r = mid - 1
else:
l = mid + 1
cnt += l
l, r = 0, len(s)-1
while l <= r:
mid = (l + r) >> 1
if s[mid] > 2*i:
r = mid - 1
else:
l = mid + 1
s = s[:l] + [2*i] + s[l:]
return cnt
|
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|
import bisect
class Solution:
def reversePairs(self, nums: List[int]) -> int:
ri, res, n = [], 0, len(nums)
for i in reversed(range(0, n)):
res += bisect.bisect_left(ri, nums[i])
bisect.insort(ri, 2 * nums[i])
return res
|
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| def reversePairs(self, nums: List[int]) -> int:
self.inversions = 0
self.Merge_Sort(nums, 0, len(nums)-1)
return self.inversions
def Merge_Sort(self, nums, l, r):
if l >= r:
return
mid = l + (r - l) // 2
self.Merge_Sort(nums, l, mid)
self.Merge_Sort(nums, mid+1, r)
i, j = l, mid+1
while i <= mid and j <= r:
if nums[j]*2 < nums[i]:
self.inversions += mid + 1 - i
j += 1
else:
i += 1
nums[l:r+1] = sorted(nums[l:r+1])
|
