325. Maximum Size Subarray Sum Equals k

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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?

解法1： O(N) HashMap

用一个hashmap存储每一个子串sum当前出现过的最早的位置（因为只要求最长的一段，所以只需要保存这个就可以了）。
然后对于每一个新球出来的sum，看是否存在sum - k存在在map中，如果存在则更新结果。要注意的是还要考虑当前的sum是否正好为k。

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