leetcode 2916. 子数组不同元素数目的平方和 II（区间更新 + 区间查询 线段树第二个板子 双闭区间 避开0）

描述

偷了一个线段树板子

不知道为啥要避开0

然后这里的更新和查找都是用双闭区间的

ac code

class SegmentTree:
    def __init__(self, n):
        self.n = n 
        self.B1 = [0]*n 
        self.B2 = [0]*n  


    def add(self, b, idx, x):
        N = self.n 
        while idx < N:
            b[idx] += x
            idx += idx & -idx

    def range_add(self, l,r,x):
        B1 = self.B1
        B2 = self.B2
        self.add(B1, l, x)
        self.add(B1, r+1, -x)
        self.add(B2, l, x*(l-1))
        self.add(B2, r+1, -x*r)

    def sum(self, b, idx):
        total = 0
        while idx > 0:
            total += b[idx]
            idx -= idx & -idx
        return total

    def prefix_sum(self, idx):
        B1 = self.B1
        B2 = self.B2
        return self.sum(B1, idx)*idx -  self.sum(B2, idx)

    def range_sum(self, l, r):
        return self.prefix_sum(r) - self.prefix_sum(l-1)

class Solution:
    def sumCounts(self, nums: List[int]) -> int:
        n = len(nums)
        seg = SegmentTree(n+5)
        MOD = 10**9+7
        seen = dict()
        ans, cur = 0, 0 
        for i, a in enumerate(nums):
            j = -1 if a not in seen else seen[a]
            s = seg.range_sum(j+2,i+1) # 双闭区间
            cur += s*2+i-j
            cur %= MOD 
            ans += cur
            ans %= MOD
            # print(i,s,cur)
            seg.range_add(j+2,i+1,1)
            seen[a] = i
        return ans

题目链接

lc2916