[leetcode/lintcode 题解] 阿里算法面试真题详解：和相同的二元子数组

描述
在由若干 0 和 1 组成的数组 A 中，有多少个和为 S 的非空子数组。

• A.length <= 30000
• 0 <= S <= A.length
• A[i] 为 0 或 1

在线评测地址：领扣题库官网

样例1
Input: A = [1,0,1,0,1], S = 2
Output: 4
Explanation: 
The 4 subarrays are bolded below:
[1,0,1]
[1,0,1]
[1,0,1,0]
[0,1,0,1]

样例2
Input: A = [0,0,0,0,0,0,1,0,0,0], S = 0
Output: 27
Explanation: 
And 27 subarrays for S.

题解
前缀和：定义一个数组sumN+1，si表示数组A中前i个元素之和，然后遍历sum数组，计算si+S(含义：前i个元素之和是si,找和为S的子数组个数)。求si+S的个数

public class Solution {
    /**
     * @param A: an array
     * @param S: the sum
     * @return: the number of non-empty subarrays
     */
    public int numSubarraysWithSum(int[] A, int S) {
        // Write your code here.
        if (A == null || A.length == 0) {
            return 0;
        }
        int prefixSum = 0;
        int res = 0;
        // counts[i] means the number of prefixSum = i
        int[] counts = new int[A.length + 1];
        counts[0] = 1;
        for (int i = 0; i < A.length; i++) {
            prefixSum += A[i];
            if (prefixSum >= S) {
                res += counts[prefixSum - S];
            }
            counts[prefixSum]++;
        }
        return res;
    }
}

更多题解参考：九章官网solution