Maximum Size Subarray Sum Equals k -- LeetCode

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.

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?

思路：O(n)算法

使用哈希表。记录下nums数组中从下标0到i之间所有数字的和。若该值为k，则更新res。若不等于，则检查该值减去k是否在哈希表中存在，若存在，则说明从之前的某一个位置到i之间的数字之和为k。我们用哈希表记录下每个和以及该和第一次出现时的i。

 class Solution {
 public:
     int maxSubArrayLen(vector<int>& nums, int k) {
         vector<int> sum(nums.size(), );
         unordered_map<int, int> help;
         int tot = , res = ;
         for (int i = , n = nums.size(); i < n; i++)
         {
             sum[i] = i ==  ? nums[] : sum[i - ] + nums[i];
             if (sum[i] == k) res = i + ;
             else if (help.count(sum[i] - k))
                 res = max(res, i - help[sum[i] - k]);
             if (help.count(sum[i]) == )
                 help.insert(make_pair(sum[i], i));
         }
         return res;
     }
 };