前缀和经典问题整理
1、一般形式 -- 区域和检索 - 数组不可变
class NumArray:def __init__(self, nums: List[int]):self.pre = [0]for num in nums:self.pre.append(self.pre[-1] + num)####或者#####self.pre = list(accumulate(nums, initial=0))def sumRange(self, left: int, right: int) -> int:return self.pre[right + 1] - self.pre[left]2、经典问题 -- 连续数组
class Solution:def findMaxLength(self, nums: List[int]) -> int:pre, m = 0, {0: -1}maxl = 0for i, num in enumerate(nums):pre += 1 if num == 1 else -1if m.get(pre, None) != None:maxl = max(i - m[pre], maxl)else:m[pre] = ireturn maxl给定一个二进制数组 nums , 找到含有相同数量的 0 和 1 的最长连续子数组，并返回该子数组的长度。
前后缀 -- 除自身以外数组的乘积
class Solution:def productExceptSelf(self, nums: List[int]) -> List[int]:right = [1] * (len(nums) + 1)for i in reversed(range(len(nums))): right[i] = right[i + 1] * nums[i]left = 1res = [1] * (len(nums))for i in range(len(nums)):res[i] = left * right[i + 1]left *= nums[i]return res前缀异或 -- 形成两个异或相等数组的三元组数目
class Solution:def countTriplets(self, arr: List[int]) -> int:n = len(arr)s = [0]for val in arr:s.append(s[-1] ^ val)cnt, total = Counter(), Counter()ans = 0for k in range(n):if s[k + 1] in cnt:ans += cnt[s[k + 1]] * k - total[s[k + 1]]cnt[s[k]] += 1total[s[k]] += kreturn ans求一个数组两两乘积之和
res, s = 0, sum(nums)
for num in nums:s -= numres += s * num
return res类似的有 -- 统计梯形的数目 I
class Solution:def countTrapezoids(self, points: List[List[int]]) -> int:m = defaultdict(int)for x, y in points:m[y] += 1res = 0arr = list(m.keys())pre = 0for i in range(len(arr)):pre = (pre + m[arr[i]] * (m[arr[i]] - 1) // 2) % (10 ** 9 + 7)for i in range(len(arr)):pre = (pre - m[arr[i]] * (m[arr[i]] - 1) // 2) % (10 ** 9 + 7) a2 = m[arr[i]] * (m[arr[i]] - 1) // 2 % (10 ** 9 + 7)res = (res + a2 * pre) % (10 ** 9 + 7)return res枚举分母，对商进行前缀求和 -- 向下取整数对和
class Solution:def sumOfFlooredPairs(self, nums: List[int]) -> int:m = Counter(nums)maxl = max(nums)pre = [0] * (maxl + 1)for i in range(1, maxl + 1):pre[i] = pre[i - 1] + m[i]res = 0for num in m:i = 1while i * num <= maxl:if maxl < (i + 1) * num - 1:res = (res + (pre[-1] - pre[i * num - 1]) * i * m[num]) % (10 ** 9 + 7)else:res = (res + (pre[(i + 1) * num - 1] - pre[i * num - 1]) * i * m[num]) % (10 ** 9 + 7)i += 1return res前缀最值 -- 有序三元组中的最大值 II
class Solution:def maximumTripletValue(self, nums: List[int]) -> int:n = len(nums)right = [0] * (n + 1)i = n - 1for num in reversed(nums):right[i] = max(right[i + 1], num)i -= 1left, maxl = 0, 0for i, num in enumerate(nums):maxl = max(maxl, (left - num) * right[i + 1])left = max(left, num)return maxl3、二维数组前缀和和差分
（1）二维数组前缀和 -- 二维区域和检索 - 矩阵不可变
代码：
class NumMatrix:def __init__(self, matrix: List[List[int]]):self.sum_matrix = [[0] * len(matrix[0]) for _ in matrix]for i in range(len(matrix)):row_sum = 0for j in range(len(matrix[i])):row_sum += matrix[i][j]self.sum_matrix[i][j] = self.sum_matrix[i - 1][j] + row_sumdef sumRegion(self, row1: int, col1: int, row2: int, col2: int) -> int:res = self.sum_matrix[row2][col2]if col1 > 0: res -= self.sum_matrix[row2][col1 - 1]if row1 > 0: res -= self.sum_matrix[row1 - 1][col2]if col1 > 0 and row1 > 0: res += self.sum_matrix[row1 - 1][col1 - 1]return res（2）二维数组差分 -- 子矩阵元素加 1
class Solution:def rangeAddQueries(self, n: int, queries: List[List[int]]) -> List[List[int]]:d = [[0] * (n + 1) for _ in range(n + 1)]for r1, c1, r2, c2 in queries:d[r1][c1] += 1d[r2 + 1][c2 + 1] += 1d[r1][c2 + 1] -= 1d[r2 + 1][c1] -= 1ans = [[0] * (n + 1) for _ in range(n + 1)]for i, row in enumerate(d[:n]):for j, x in enumerate(row[:n]):ans[i + 1][j + 1] = ans[i + 1][j] + ans[i][j + 1] - ans[i][j] + xdel ans[0]for row in ans:del row[0]return ans数组差分可以看成函数微分，数组前缀和可以看成函数积分，所以差分数组的前缀和就是原数组
4、字符串哈希 + 前缀和 -- 不同的循环子字符串
class Solution:def distinctEchoSubstrings(self, text: str) -> int:n = len(text)mod, base = 10**9 + 7, 31pre, mul = [0] * (n + 1), [1] + [0] * nfor i in range(1, n + 1):pre[i] = (pre[i - 1] * base + ord(text[i - 1])) % modmul[i] = mul[i - 1] * base % moddef get_hash(l, r):return (pre[r + 1] - pre[l] * mul[r - l + 1] % mod + mod) % modseen = {x: set() for x in range(n)}ans = 0for i in range(n):for j in range(i + 1, n):l = j - iif j + l <= n:hash_left = get_hash(i, j - 1) if hash_left not in seen[l - 1] and hash_left == get_hash(j, j + l - 1):ans += 1seen[l - 1].add(hash_left)return ans5、进阶问题
（1）个数前缀和 -- 查询差绝对值的最小值
class Solution:def minDifference(self, nums: List[int], queries: List[List[int]]) -> List[int]:pre = [[0] for _ in range(101)]for n in nums:for i in range(101):if i == n:pre[i].append(pre[i][-1] + 1)else:pre[i].append(pre[i][-1])res = []for f, t in queries:last, minl = None, inffor i in range(101):if pre[i][t + 1] - pre[i][f] > 0:if last is not None:minl = min(i - last, minl)last = ires.append(minl if minl != inf else -1)return res（2）统计回文子序列数目
class Solution:def countPalindromes(self, s: str) -> int:suf = [0] * 10suf2 = [0] * 100for d in map(int, reversed(s)):for j, c in enumerate(suf):suf2[d * 10 + j] += csuf[d] += 1ans = 0pre = [0] * 10pre2 = [0] * 100for d in map(int, s):suf[d] -= 1for j, c in enumerate(suf):suf2[d * 10 + j] -= c # 撤销ans += sum(c1 * c2 for c1, c2 in zip(pre2, suf2)) # 枚举所有字符组合for j, c in enumerate(pre):pre2[d * 10 + j] += cpre[d] += 1return ans % (10 ** 9 + 7)（3）统计上升四元组
class Solution:def countQuadruplets(self, nums: List[int]) -> int:n = len(nums)more = [[0] * n for _ in range(n + 1)]less = [[0] * n for _ in range(n + 1)]for j in reversed(range(n)):for k in reversed(range(j + 1, n)):if nums[j] < nums[k]:more[k][j] = more[k + 1][j] + 1else:more[k][j] = more[k + 1][j]for k in range(n):for j in range(k):if nums[k] > nums[j]:less[j][k] = less[j - 1][k] + 1else:less[j][k] = less[j - 1][k]res = 0for k in range(n):for j in range(k):if nums[k] < nums[j]:res += less[j][k] * more[k][j]return res(4)前缀和 + 哈希 + 同余 -- 统计美丽子字符串 II
class Solution:def beautifulSubstrings(self, s: str, k: int) -> int:k = self.sqrt(k * 4)cnt = Counter([(0, 0)]) ans = pre_sum = 0for i, c in enumerate(s):pre_sum += 1 if c in "aeiou" else -1p = ((i + 1) % k, pre_sum)ans += cnt[p]cnt[p] += 1return ansdef sqrt(self, n: int) -> int:res = 1i = 2while i * i <= n:i2 = i * iwhile n % i2 == 0:res *= in //= i2if n % i == 0:res *= in //= ii += 1if n > 1:res *= nreturn res（5）二维矩阵压缩到一维 + 前缀和 + 哈希 -- 矩形区域不超过 K 的最大数值和
class Solution:def numSubmatrixSumTarget(self, matrix: List[List[int]], target: int) -> int:m, n = len(matrix), len(matrix[0])res = 0for i in range(1, n + 1):presum = [0] * (m + 1)for j in range(i, n + 1):a = 0d = defaultdict(int, {0:1})for fixed in range(1, m + 1):presum[fixed] += matrix[fixed-1][j-1]a += presum[fixed]res += d[a - target]d[a] += 1return res