leetcode974. Subarray Sums Divisible by K 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...
所以我们简历Map存储累计和sum出现的次数,如果sum-k存在于map中,则累加和为k的子数组一定存在。 publicintsubarraySum(int[] nums,intk){ HashMap<Integer, Integer> map =newHashMap<>(); map.put(0,1);//初始加入0,例如nums【0】=10,k=10,如果0不在map中就少一个resintres=0;intcur=0;for(inti=...
如果sum比k小,那么就需要扩大窗口,right右移(注意这里可能会到达最右端的判断);如果sum==k,满足条件,窗口缩小,left右移;如果sum>k,窗口缩小,left右移。 整理代码如下: 1publicintsubarraySum(int[] nums,intk) {2intleft = 0;3intright = 0;4intsum = nums[0];5intcount = 0;6while( right < nu...
Given an array of integersnumsand an integerk, returnthe total number of subarrays whose sum equals tok. A subarray is a contiguousnon-emptysequence of elements within an array. Example 1: Input: nums = [1,1,1], k = 2 Output: 2 Example 2: Input: nums = [1,2,3], k = 3 Outp...
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Leetcode 560. Subarray Sum Equals K Subarray Sum Equals K 题目大意:给定一个数组,要求找到不同的连续子序列,他们的和为k,问这样的连续子序列有多少个 题目思路:首先根据原始的想法,我们可以想到连续子序列的问题可以通过前缀和的方式进行解决,可以通过n^2的问题解决,但是这个做法并不太优,所以我们...
Given an array nums and a target value k, find the maximum length of a subarray that sums to k. If there isn't one, return 0 instead. Note:The sum of the entire nums array is guaranteed to fit within the 32-bit signed integer range. Example 1: Given nums = [1, -1, 5, -2...
LeetCode Top 100 Liked Questions 560. Subarray Sum Equals K (Java版; Medium) 题目描述 AI检测代码解析 Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k.
[Leetcode] 560. Subarray Sum Equals K Problem: https://leetcode.com/problems/subarray-sum-equals-k/ Solution1: 利用累计和 Time complexity: O(n2) Space complexity: O(n) Solution2: 移动subarray的起点和终点,计算两点之间的和 Time complexity: O(n2) Space complexit......
classSolution {public:intmaxSubArrayLen(vector<int>& nums,intk) {if(nums.empty())return0;intres =0; unordered_map<int, vector<int>>m; m[nums[0]].push_back(0); vector<int> sum =nums;for(inti =1; i < nums.size(); ++i) { ...