Given an array nums and a target value k, find the maximum length of a subarray that sums to k. If there isn't one, return 0 instead.
Note:
The sum of the entire nums array is guaranteed to fit within the 32-bit signed integer range.

Example 1:
Given nums = [1, -1, 5, -2, 3], k = 3,
return 4. (because the subarray [1, -1, 5, -2] sums to 3 and is the longest)
Example 2:
Given nums = [-2, -1, 2, 1], k = 1,
return 2. (because the subarray [-1, 2] sums to 1 and is the longest)

Follow Up:
Can you do it in O(n) time?

Solution：Hashmap

思路：累積sum，存入hashmap(sum_value, index)，已有的話不更新(為保證最長(zhǎng))，并在hashmap中找是否有另一半作差能得到k(subarray和為k)
Time Complexity: O(N) Space Complexity: O(N)

Solution Code：

class Solution {
    public int maxSubArrayLen(int[] nums, int k) {
        int sum = 0, max = 0;
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        map.put(0, -1); // corner case for from 0 .. i sum
        for (int i = 0; i < nums.length; i++) {
            sum = sum + nums[i];
            if (map.containsKey(sum - k)) max = Math.max(max, i - map.get(sum - k));
            if (!map.containsKey(sum)) map.put(sum, i);
        }
        return max;
    }
}