Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k.

Example 1:

1
2
Input:nums = [1,1,1], k = 2
Output: 2

Note:

  1. The length of the array is in range [1, 20,000].
  2. The range of numbers in the array is [-1000, 1000] and the range of the integer k is [-1e7, 1e7].

因为数组中可能出现负数,所以不能用双指针。

用一个哈希表记录所有前缀和出现的下标,然后对于每一个下标i对应的前缀和sum,找到sum+k出现的下标,其中大于等于i的下标的个数就是从i开始的和为k的连续子串的个数。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
class Solution {
public:
int subarraySum(vector<int>& nums, int k) {
unordered_map<int, vector<int>> sum2index;
int sum = 0, ans = 0;
for (int i = 0; i < nums.size(); i++) {
sum += nums[i];
sum2index[sum].push_back(i);
}
sum = 0;
for (int i = 0; i < nums.size(); i++) {
if (!sum2index.count(sum + k)) {
sum += nums[i];
continue;
}
vector<int> &indexes = sum2index[sum + k];
int j;
for (j = 0; j < indexes.size() && indexes[j] < i; j++);
ans += (indexes.size() - j);
sum += nums[i];
}
return ans;
}
};