LeetCode-Binary Subarrays With Sum
阿新 • • 發佈:2018-12-18
Description: In an array A of 0s and 1s, how many non-empty subarrays have sum S?
Example 1:
Input: A = [1,0,1,0,1], S = 2
Output: 4
Explanation:
The 4 subarrays are bolded below:
[1,0,1,0,1]
[1,0,1,0,1]
[1,0,1,0,1]
[1,0,1,0,1]
Note:
- A.length <= 30000
- 0 <= S <= A.length
- A[i] is either 0 or 1.
題意:給定一個只包含數字0和1的陣列A, 和一個整數S,計算陣列A的所有子陣列中滿足1的個數為S的數量;
解法:定義一個新的陣列sumOne,用來儲存陣列A中當前下標位置處1的個數(即從下標0開始到當前下標位置處1的個數);這樣,我們只要找到sumOne中所有滿足兩個位置處的1的個數差為S的情況即可; 例如:對於陣列A = [1,0,1,0,1],我們可以得到陣列sumOne = [1,1,2,2,3] ;
Java
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[i] = A[i] == 1 ? 1 + sumOne[i - 1] : sumOne[i - 1]; } while (st < A.length) { ed = st; while (ed < A.length) { int sum = st == 0 ? sumOne[ed] : sumOne[ed] - sumOne[st - 1]; if (sum > S) break; result = sum == S ? result + 1 : result; ed++; } st++; } return result; } }