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。
1 class Solution { 2 public: 3 int maxSubArrayLen(vector<int>& nums, int k) { 4 vector<int> sum(nums.size(), 0); 5 unordered_map<int, int> help; 6 int tot = 0, res = 0; 7 for (int i = 0, n = nums.size(); i < n; i++) 8 { 9 sum[i] = i == 0 ? nums[0] : sum[i - 1] + nums[i]; 10 if (sum[i] == k) res = i + 1; 11 else if (help.count(sum[i] - k)) 12 res = max(res, i - help[sum[i] - k]); 13 if (help.count(sum[i]) == 0) 14 help.insert(make_pair(sum[i], i)); 15 } 16 return res; 17 } 18 };
Maximum Size Subarray Sum Equals k -- LeetCode
原文：http://www.cnblogs.com/fenshen371/p/5154495.html