与Zero Sum Subarray 题的变化之处有两个地方,第一个是判断是否存在哈希表中时需要使用hash.find(curr_sum - k), 最终返回结果使用result.push_back(hash[curr_sum - k]);而不是result.push_back(hash[curr_sum]); 复杂度分析 略,见Zero Sum Subarray | Data Struc
Can you solve this real interview question? Subarray Sum Equals K - Given an array of integers nums and an integer k, return the total number of subarrays whose sum equals to k. A subarray is a contiguous non-empty sequence of elements within an array.
HashSet<Integer> sums=newHashSet<>(); intsum=0; for(inti=0;i<nums.length;++i){ sum+=nums[i]; if((k!=0&&sums.contains(sum%k))||(i!=0&∑==k)){ returntrue; }elseif(k!=0){ sums.add(sum%k); } } returnfalse; } } 方法三:用HashMap保存sum对k取余数,如果前序有余数也为...
注意上图中prefixSum的第一个数0,是为了解决当i == 0的时候的subarray(i, j)的和,这时对应的prefixSum[i - 1]应该是0 有了prefixSum array,这个问题就转化成了 for each j:how many i < j satisfies prefixSum[i] = prefixSum[j] - k 这意味着对于每个j,我们需要记录之前所有的prefixSum,然后在这...
Basic idea, for array starting at every A[i], find the shortest one with sum at leat K. In my solution,for B[i], find the smallest j that B[j] - B[i] >= K. Keep this in mind for understanding two while loops. What is the purpose of first while loop?
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给一个都是整数的数组,返回 和可以被K整除的子数组 数。Given an array A of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by K.Example 1:Input: A = [4,5,0,-2,-3,1], K = 5Output: 7Explanation: There are 7 subarrays with a sum ...
遍历数组,累加到每一个元素的sum添加到map中,得到0, sum0,sum1,sum2, …., sum(n-1).如果sumj - sumi = k,即可得sum(i+1…j)为k,即计数+1。 class Solution public int subarraySum(int[] nums, int k) { if (nums == null || nums.length == 0) { ...
class Solution { public: int subarraySum(vector<int>& nums, int k) { int n = nums.size(), sum = 0, res = 0; vector<int> pre(n); unordered_map<int, int> mp; mp[0] = 1; for(int i = 0; i < n; ++i) { sum += nums[i]; pre[i] = sum; } for(int i = 0; i...
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: