题目
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?

答案
Idea

First pass
Calculate sum up to nums[i], store in a map which uses sum as key, i as value.
Second pass
for each nums[i], look up if there is an index j such that nums[i] + .... + nums[j] = k
which means sum up to index j(inclusive) - sum up to index i(exclusive) = k
if the map contains x = k + sum up to index i (exclusive), we know subarray from index i to index map.get(x) is the maximum size subarray that sums to k

public int maxSubArrayLen(int[] nums, int k) {
        Map<Integer, Integer> map = new HashMap<>();
        int sum = 0;
        // First pass
        for(int i = 0; i < nums.length; i++) {
            sum += nums[i];
            map.put(sum, i);
        }

        int max = 0;
        sum = 0;
        // Second pass
        for(int i = 0; i < nums.length; i++) {
            // already have need nums[i] + nums[i+1] + .... + nums[j] = k
            // sum up to j - previous sum
            Integer lookup = map.get(k + sum);
            if(lookup != null) {
                max = Math.max(lookup - i + 1, max);
            }
            sum += nums[i];
        }
        return max;
    }