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?

一刷
题解：我们把所有从0到i的sum存到map中，map.put(sum, i+1), 并且如果有重复的sum, 不更新，保证map中的value最小。
这样的话，如果当前的sum!=k, 但是sum-k在map中，表示有一段subarray可以等于k, 于是max = Math.max(max, i+1-map.get(sum-k));, 这就是map不更新，保证map保存最短的subarray

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