python刷题总结_【python刷题】差分数组

前缀和主要适用的场景是原始数组不会被修改的情况下，频繁查询某个区间的累加和。

差分数组的主要适用场景是频繁对原始数组的某个区间的元素进行增减。

class Difference:

def __init__(self, nums):

self.nums = nums

self.diff = self.make_diff(self.nums)

def make_diff(self, nums):

diff = [0 for _ in range(len(nums))]

diff[0] = nums[0]

for i in range(1,len(nums)):

diff[i] = nums[i] – nums[i-1]

return diff

# 给闭区间[i,j]增加val

# 原理很简单，回想diff数组反推nums数组的过程，diff[i] += 3意味着给nums[i..]所有的元素都加了 3，然后diff[j+1] -= 3又意味着对于nums[j+1..]所有元素再减 3，那综合起来，是不是就是对nums[i..j]中的所有元素都加 3 了？

def increment(self, i, j, val):

self.diff[i] += val

if j + 1 < len(self.diff):

self.diff[j + 1] -= val

def recover(self):

rdiff = [0 for _ in range(len(self.diff))]

rdiff[0] = self.diff[0]

for i in range(1,len(self.nums)):

rdiff[i] = rdiff[i-1] + self.diff[i]

return rdiff

difference = Difference([8,5,9,6,1])

print(difference.diff)

rdiff = difference.recover()

print(rdiff)

difference.increment(2,3,1)

print(difference.diff)

rdiff = difference.recover()

print(rdiff)

结果：

[8, -3, 4, -3, -5]

[8, 5, 9, 6, 1]

[8, -3, 5, -3, -6]

[8, 5, 10, 7, 1]

不妨去试试力扣第 1109 题「航班预订统计]。

labuladong的算法小抄