LeetCode 325. Maximum Size Subarray Sum Equals k
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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:
1. 最简单粗暴的做法是,求和所有subarray并选出和为k的最长subarray,这样的复杂度是:列举所有subarray有o(n^2)个,然后每个subarray求和复杂度是o(n),总复杂度o(n^3).
2. 不粗暴的做法呢?不用计算所有subarray的和,只需计算前i个连续数的和
3. 需要提取利用的信息,就是前i个连续数的和
4. 本题可以看出利用hashtable可以大大降低复杂度!!!
int maxSubArrayLen(vector<int>& nums, int k) { unordered_map<int,int> mm; mm[0]=-1; int sum=0,res=0; for(int i=0;i<nums.size();i++){ sum+=nums[i]; if(mm.count(sum-k)){ res=max(res,i-mm[sum-k]); } if(mm.count(sum)==0) mm[sum]=i; } return res;}
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