README_EN.md

560. Subarray Sum Equals K

中文文档

Description

Given an array of integers nums and an integer k, return the total number of continuous subarrays whose sum equals to k.

 

Example 1:

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

Example 2:

Input: nums = [1,2,3], k = 3
Output: 2

 

Constraints:

• 1 <= nums.length <= 2 * 104
• -1000 <= nums[i] <= 1000
• -107 <= k <= 107

Solutions

Python3

class Solution:
    def subarraySum(self, nums: List[int], k: int) -> int:
        counter = Counter({0: 1})
        ans = s = 0
        for num in nums:
            s += num
            ans += counter[s - k]
            counter[s] += 1
        return ans

Java

class Solution {
    public int subarraySum(int[] nums, int k) {
        Map<Integer, Integer> counter = new HashMap<>();
        counter.put(0, 1);
        int ans = 0, s = 0;
        for (int num : nums) {
            s += num;
            ans += counter.getOrDefault(s - k, 0);
            counter.put(s, counter.getOrDefault(s, 0) + 1);
        }
        return ans;
    }
}

TypeScript

function subarraySum(nums: number[], k: number): number {
    let ans = 0,
        s = 0;
    let counter = new Map();
    counter[0] = 1;
    for (const num of nums) {
        s += num;
        ans += counter[s - k] || 0;
        counter[s] = (counter[s] || 0) + 1;
    }
    return ans;
}

C++

class Solution {
public:
    int subarraySum(vector<int>& nums, int k) {
        unordered_map<int, int> counter;
        counter[0] = 1;
        int ans = 0, s = 0;
        for (int& num : nums)
        {
            s += num;
            ans += counter[s - k];
            ++counter[s];
        }
        return ans;
    }
};

Go

func subarraySum(nums []int, k int) int {
	counter := map[int]int{0: 1}
	ans, s := 0, 0
	for _, num := range nums {
		s += num
		ans += counter[s-k]
		counter[s]++
	}
	return ans
}

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