560. Subarray Sum Equals K

Description

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].

Problem URL

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Solution

给一个数组，找到数组中有多少个连续的子序列的和为k。

1.Brute force
Use two for loop, calculate from I to rear of nums[], if some sum is k, count++;

2. Pre sum
   sum[I, j] = sum[0, j] - sum[0, I]. So we could calculate pre sums, and store them in a Hash map. We put (0, 1) in the hash map for initial value. It means 1 frequence of sum = 0. Then iterate nums, for every num in numbers, if (curSum - k) is in preSum, it means cur - pre = k. So we could get the frequence by preSum.get(curSum - k).

For every calculated curSum, we should store it in preSum, if it is in, plus one; if not, 1. We could use getOrDefault() of hash map to realize it.

Code

1.Brute Force

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

2. Pre sum

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

Time Complexity: O(n^2) O(n)
Space Complexity: O(1) O(n)

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