publicstaticintnumberOfSubarrays(int[]arr,int target){HashMap<Integer,Integer>map=newHashMap<>();int pre=0;map.put(0,1);int cnt=0;for(int i=0;i<arr.length;i++){pre+=arr[i];if(map.containsKey(pre-target)){cnt+=map.get(pre-target);}map.put(pre,map.getOrDefault(pre,0)+1)...
public:intnumSubarraysWithSum(vector<int>& A,intS) {intres =0,sum=0, left =0, n = A.size();for(inti =0; i < n; ++i) {sum+= A[i];while(left < i &∑> S)sum-= A[left++];if(sum< S)continue;if(sum== S) ++res;for(intj = left; j < i && A[j] ==0; ++j)...
1classSolution {2publicintnumSubarraysWithSum(int[] A,intS) {3if(A ==null|| A.length == 0){4return0;5}67int[] count =newint[A.length+1];8count[0] = 1;9intsum = 0;10intres = 0;11for(inta : A){12sum +=a;13if(sum >=S){14res += count[sum-S];15}1617count[sum]...
class Solution { public int numSubarraysWithSum(int[] A, int S) { if (A.length == 0) return 0; int result = 0; int[] sumOne = new int[A.length]; int st = 0; int ed = 0; sumOne[0] = A[0] == 1 ? 1 : 0; for (int i = 1; i < A.length; i++) { sumOne[...
classSolution{publicintnumSubarraysWithSum(int[] A,intS){int[] ps =newint[A.length +1]; ps[0] =1;intsum=0;intret=0;for(intv: A) { sum += v;if(sum - S >=0) { ret += ps[sum - S]; } ps[sum] +=1; }returnret; ...
题目地址:https://leetcode.com/problems/binary-subarrays-with-sum/description/ 题目描述 In an arrayAof0s and1s, how manynon-emptysubarrays have sumS? Example 1: Input: A = [1,0,1,0,1], S =2Output:4Explanation: The4subarrays are bolded below: ...
publicintnumSubarraysWithSum(int[] A,intS) {intn = A.length, res =0,sum=0;int[]map= newint[n +1];map[0] =1;for(inti =0; i < n; ++i) {sum+= A[i];if(sum>= S) res +=map[sum- S]; ++map[sum]; }returnres; ...
1classSolution {2func numSubarraysWithSum(_ A: [Int], _ S: Int) ->Int {3varn:Int =A.count4varcum:[Int] = [Int](repeating:0,count: n +1)5foriin0..<n6{7cum[i +1] = cum[i] +A[i]8}910varret:Int =011varf:[Int] = [Int](repeating:0,count:30003)12foriin0...n13...