Given an integer array, find a subarray where the sum of numbers iszero. Your code should return the index of the first number and the index of the last number. Notice: There is at least one subarray that it's
class Solution { public boolean checkSubarraySum(int[] nums, int k) { int acc = 0; Map<Integer, Integer> map = new HashMap<>(); map.put(0, -1); for (int i = 0; i < nums.length; i++){ acc += nums[i]; int rem = acc % k; if (map.containsKey(rem)){ if (i - m...
FindTabBarSize numsand an integerthe number of non-emptysubarrayswith a sumgoal. Asubarrayis a contiguous part of the array. Example 1: Input:nums = [1,0,1,0,1], goal = 2Output:4Explanation:The 4 subarrays are bolded and underlined below: [1,0,1,0,1] [1,0,1,0,1] [1,0,...
第一种是prefix sum + 双指针。 例如Minimum Size Subarray Sum - LeetCode: Given an array of n positive integers and a positive integers, find the minimal length of a contiguous subarray of which the sum ≥s. If there isn't one, return 0 instead. 用O(N)时间得到一个prefix sum array,可...
class Solution(object): def numSubarraysWithSum(self, A, S): """ :type A: List[int] :type S: int :rtype: int """ N = len(A) res = 0 preS = 0 count = collections.Counter({0 : 1}) for i in range(1, N + 1): preS += A[i - 1] res += count[preS - S] count...
此时还需要考虑一种情况,就是当窗口左边有连续0的时候,因为0并不影响 sum,但是却要算作不同的子数组,所以要统计左起连续0的个数,并且加到结果 res 中即可,参见代码如下: 解法二: classSolution{ public:intnumSubarraysWithSum(vector<int>& A,intS) {intres =0,sum=0, left =0, n = A.size();...
www.lintcode.com/en/problem/minimum-size-subarray-sum/ 【题目解析】 sum为前i个数的和,长度比nums多一个,sum[0] = 0。这样从0开始一直到len,遍历sum计算sum[j]与sum[i]的差大于等于s的时候的j-i长度,把它与minlen比较,如果比minlen小就更新minlen。
the paragraph below was copied from his paper (with a little modifications) algorithm that operates on arrays: it starts at the left end (element A[1]) and scans through to the right end (element A[n]), keeping track of the maximum sum subvector seen so far. The maximum is initially...
Find sum of all subarray sums out of an array.Example:Input array: [1, 2, 3, 4] Output: 50 Solution:Of course, there exists an easy solution where we can use three for loops with time complexity (O (n3)). The outer loop and intermediate loop are to iterate through all subarrays ...
2. Solution **解析:**Version 1,简单粗暴,前i个元素总和大于0,则这些元素对总和是有贡献的,要保留,否则,则丢弃前i个元素。重新开始执行上述操作,每次加完记得更新最大值。Version 2,采用动态规划求解,首先定义状态,dp[i]是以nums[i]为结尾的连续子数组的最大和,状态转移方程为:如果dp[i-1]>0,无论nums...