樹狀陣列

更高效地對點、區間進行查詢、更新

ref:樹狀陣列詳解

k為i的二進位制中從低位到高位的連續零長度

C[i] = A[i-2^k+1] +A[i-2^k+2] + ... + A[i]

2^k = i&(-i)

sum[i] = C[i] + C[i - 2^k1] +C[i - 2^k1 - 2^k2]

eg：315. 計算右側小於當前元素的個數

def lowBit(x:int) -> int:
    return x&(-x)

def update(x: int, n: int, y: int, c:[]):
     i = x
     while
 
 i < n:
         c[i] += y
         i += lowBit(i)
def getSum(x: int, c:[]) -> int:
    ans = 0
    i = x
    while i > 0:
        ans += c[i]
        i -= lowBit(i)
    return ans
class Solution:
    def countSmaller(self, nums:[int]) ->[int]:
        buckets = set()
        for num in nums:
            buckets.add(num)
        buckets 
 
= sorted(buckets)
        pos = {}
        for i, e in enumerate(list(buckets)):
            pos[e] = i
        n = len(nums)
        counts = [0] * n
        c = [0] * n
        for i in range(n - 1, -1, -1):
            id = pos[nums[i]] + 1
            update(id, n, 1, c)
            counts[i] = getSum(id - 1, c)
        
 
return counts