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?
解题思路：1、累加nums[i]
2、查找hashmap有没有sum - k
3、如果当前sum没有在hashmap里，保存当前sum的位置
代码如下：
public class Solution { public int maxSubArrayLen(int[] nums, int k) { HashMap<Integer,Integer> hm = new HashMap<Integer,Integer>(); int sum = 0, max = 0; for (int i = 0; i < nums.length; i ++) { sum += nums[i]; if (sum == k) { max = i + 1; } else if (hm.containsKey(sum - k)) { max = Math.max(max, i - hm.get(sum - k)); } if (!hm.containsKey(sum)) { hm.put(sum, i); } } return max; } }