[leetcode] 325. Maximum Size Subarray Sum Equals k 解题报告

题目链接:https://leetcode.com/problems/maximum-size-subarray-sum-equals-k/

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: 可以用hash表来保存前n个数的和,然后每次去查是否当前和与目标值之差已经存在, 是的话说明我们找到了一个序列,然后更新最大长度大小. 哦, 还有就是如果有相同的和, 那就不管了, 因为我们要的最长的子串, 肯定是留着前面的一个值更优.

代码如下:

class Solution {public:    int maxSubArrayLen(vector<int>& nums, int k) {        if(nums.size()==0) return 0;        unordered_map<int, int> hash;        int sum = 0, Max = 0;        hash[0] = 0;        for(int i = 0; i < nums.size(); i++)        {            sum += nums[i];            if(hash.count(sum-k)) Max = max(Max, i-hash[sum-k]+1);            if(!hash.count(sum)) hash[sum] = i+1;        }        return Max;    }};