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;
}
};