class Solution: def minSubArrayLen(self, target: int, nums: List[int]) -> int: sub_len = 0 min_len = float('inf') l = 0 sn = 0 for i in range(len(nums)): sub_len += 1 sn += nums[i] if sn < target: continue while sn - nums[l] >= target: sub_len -= 1 sn -...
【原题】 Given a list of non-negative numbers and a target integer k, write a function to check if the array has a continuous subarray of size at least 2 that sums up to the multiple of k, that is, sums up to n*k where n is also an integer. Example 1: Input: [23, 2, 4,...
代码(Python3) class Solution: def maximumSubarraySum(self, nums: List[int], k: int) -> int: # ans 维护所有长度为 k 且数字各不相同的子数组中,子数组和的最大值 ans: int = 0 # sum 维护当前滑动窗口 [l, r] 内的数字和 sum: int = 0 # num_to_cnt 表示滑动窗口 [l, r] 内每个数...
LeetCode LintCode和大于S的最小子数组Minimum Size Subarray Sum题目分析 题目给定一个由 n 个整数组成的数组和一个正整数 s ,请找出该数组中满足其和 ≥ s 的最小长度子数组。如果无解,则返回 -1。...样例 给定数组 [2,3,1,2,4,3] 和 s = 7, 子数组 [4,3] 是该条件下的最小长度子数组。分...
Given an array A of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by K. Example 1: Note: 给定一个集合,求这个集合中子集的个数,其中对子集的要求是子集中元素的和能被k整除。 记数组pre[i+1]表示前 i 位元素的和mo...7...
907. Sum of Subarray Minimums Given an array of integers A, find the sum of min(B), where B ranges over every (contiguous) subarray of A. Since the answer may be large, return the answer modulo 10^9 + 7....
907. Sum of Subarray Minimums Given an array of integers A, find the sum of min(B), where B ranges over every (contiguous) subarray of A. Since the answer may be large, return the answer modulo 10^9 + 7.Example 1: Input: [3,1,2,4] Output: 17 Explanation: Subarrays are [3]...
链接:https://leetcode-cn.com/problems/minimum-size-subarray-sum python # 长度最小的子数组 # 找出该数组中满足其和 ≥ target 的长度最小的 连续子数组[numsl, numsl+1, ..., numsr-1, numsr] ,并返回其长度。如果不存在符合条件的子数组,返回 0 。
代码二 (python) 用了二分法来缩小空间,(腾讯面试被问到,当时觉得自己没写好,所以结束后重新写一下,发现效率没有双指针高) class Solution: def minSubArrayLen(self, s: int, nums: List[int]) -> int: n=len(nums) sums=[0]*(n+1) for i in range(1,n+1): ...
Learn how to check the maximum sum of all stacks after popping some elements from them using Python with this detailed guide.