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] #...
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...
当前标签:PrefixSum [CodeForces] The Number Of Good Substrings Review->Improve 2022-01-17 13:44阅读:56评论:0推荐:0 [LeetCode 1712] Ways to Split Array Into Three Subarrays Review->Improve 2021-01-13 05:21阅读:492评论:0推荐:0 [LeetCode 1674] Minimum Moves to Make Array Complementary...
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。
LeetCode - 303. 区域和检索 - 数组不可变 给定一个整数数组nums,处理以下类型的多个查询: 计算索引left和right(包含left和right)之间的nums元素的和,其中left <= right 实现NumArray类: NumArray(int[] nums)使用数组nums初始化对象 int sumRange(int left, int right)返回数组nums中索引left和right之间的元素...
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.1300.Sum of Mutated Array Closest to Target 转变数组后最接近目标值的数组和 双指针37 困难70 LeetCode223 数组64 Medium80 每日一题131 二分搜索12 前缀和8 2020-06-03 » LeetCode.837.New 21 Game 新21点 困难70 LeetCode223 数组64 Medium80 每日一题131 动态规划25 前缀和8 概率1...
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/** * Source : https://oj.leetcode.com/problems/longest-common-prefix/ * * Created by lverpeng on 2017/7/10. * * Write a function to find the longest common prefix string amongst an array of strings. */ public class LongestCommonPrefix { /** * 依次比较每个字符串的每个字符是否相同...