# 560.Subarray-Sum-Equals-K

## 560. Subarray Sum Equals K

## 题目地址

<https://leetcode.com/problems/subarray-sum-equals-k/>

## 题目描述

```
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:
Input:nums = [1,1,1], k = 2
Output: 2
Note:
The length of the array is in range [1, 20,000].
The range of numbers in the array is [-1000, 1000] and the range of the integer k is [-1e7, 1e7].
```

## 代码

### Approach #2 Commulative Sum

```java
class Solution {
  public int subarraySum(int[] nums, int k) {
    int count = 0;
    int[] sum = new int[nums.length + 1];
    sum[0] = 0;
    for (int i = 1; i <= nums.length; i++) {
      sum[i] = sum[i - 1] + nums[i - 1];
    }

    for (int start = 0; start < nums.length; start++) {
      for (int end = start + 1; end <= nums.length; end++) {
        if (sum[end] - start[start] == k) {
          count++;
        }
      }
    }

    return count;
  }
}
```

### Approach #3 Without space

```java
class Solution {
  public int subarraySum(int nums, int k) {
    int count = 0;
    for (int start = 0; start < nums.length; start++) {
      int sum = 0;
      for (int end = start; end < nums.length; end++) {
        sum += nums[end];
        if (sum == k) {
          count++;
        }
      }
    }
    return count;
  }
}
```

### Approach #4 Hashmap

**Complexity Analysis**

* Time complexity : O(n)
* Space complexity : O(n)

```java
class Solution {
  public int subarraySum(int[] nums, int k) {
    int count = 0;
    int sum = 0;
    HashMap<Integer, Integer> map = new HashMap();
    map.put(0, 1);
    for (int i = 0; i < nums.length; i++) {
      sum += nums[i];
      if (map.containsKey(sum - k)) {
        count += map.get(sum - k);
      }
      map.put(sum, map.getOrDefault(sum, 0) + 1);
    }

    return count;
  }
}
```

### Approach #1 Brute Force \[Time Limit Exceeded]

```java
class Solution {
  public int subarraySum(int[] nums, int k) {
        int count = 0;
    for (int start = 0; start < nums.length; start++) {
      for (int end = start + 1; end <= nums.length; end++) {
        int sum = 0;
        for (int i = start; i < end; i++) {
          sum += nums[i];
        }
        if (sum == k) {
          count++;
        }
      }
    }

    return count;
  }
}
```


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