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prefixSum[1] = prefixSum[0] + arr[1] = 30, prefixSum[2] = prefixSum[1] + arr[2] = 40 and so on. 其python实现为: # prefix sum array的实现 def fillPrefixSum(arr,n,prefixSum): prefixSum[0] = arr[0] for i in range(1,n): prefixSum[i] = prefixSum[i-1] + arr[i] #...
显然a1和a5之间的区间值 = sum5 - sum1 a3和a7之间的区间值 = sum7 - sum3 现在有一道LeetCode上面的算法题是这样的 => Subarray Sum Equals K 也就是说给一个数组 A = [a1, a2, a3, a4, a5, a6, a7, a8], 以及一个整数K, 计算差值为K的两个元素的区间数 比如整个数组A中有 (a1+a2+a3...
2. 参考Prefix Sum & Dictionary Time and Space O(n) for -- Find a number of continuous subarrays/submatrices/tree paths that sum to targetT: O(n), S: O(n) classSolution:defsubarraySum(self, nums: List[int], target: int) ->int: count, curSum, d=0, 0, collections.Counter()forn...
0167-two-sum-ii-input-array-is-sorted.rs 0169-majority-element.rs 0179-largest-number.rs 0190-reverse-bits.rs 0191-number-of-1-bits.rs 0198-house-robber.rs 0199-binary-tree-right-side-view.rs 0200-number-of-islands.rs 0202-happy-number.rs 0205-isomorphic-strings.rs 0206-reverse-linked-...
地方可以用到,只要题目是在一个subarray里面找出一个target(无论是出现的次数,还是最长或者最短subarray)基本都可以用HashMap + prefixSum 来做,这https://leetcode.com/problems/minimum-operations-to-reduce-x-to-zero/discuss/935935/Java-Detailed-Explanation-O(N)-Prefix-SumMap-Longest-Target-Sub-Array...
有一个书店老板,他的书店开了n分钟。每分钟都有一些顾客进入这家商店。给定一个长度为n的整数数组customers,其中customers[i]是在第i分钟开始时进入商店的顾客数量,所有这些顾客在第i分钟结束后离开。 在某些分钟内,书店老板会生气。 如果书店老板在第i分钟生气,那么grumpy[i] = 1,否则grumpy[i] = 0。
代码 遍历+比较最大值+时间O(n) Python3 解题思路 此处撰写解题思路 代码 classSolution:defnumTimesAllBlue(self,light:List[int])->int:sum1=0max1=0foriinrange(len(light)):max1=max(max1,light[i])ifi+1==max1:sum1+=1returnsum1
LeetCode 1018. Binary Prefix Divisible By 5 Given an array A of 0s and 1s, consider N_i: the i-th subarray from A[0] to A[i] interpreted as a binary number (from most-significant-bit to least-significant-bit.) Return a list of booleans answer, ...Leetcode 1018. Binary Prefix...
你将随机生成一个长度为 \(k\) 的数组 \(a\),其中 \(a_i\) 有 \(p_i\) 概率为 \(1\),否则为 \(-1\)。定义其前缀和数组 \(s_i = \sum\limits_{j = 1}^{i}a_j, i \in [0, k]\)。如果前缀和数组的最大值为 \(t(t \in [0, k])\) 那么你将获得 \(h_t\) 的权值。