原题链接在这里：https://leetcode.com/problems/maximum-size-subarray-sum-equals-k/
题目：
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.
Note:
The sum of the entire nums array is guaranteed to fit within the 32-bit signed integer range.
Example 1:
Input: nums =[1, -1, 5, -2, 3], k =3
Output: 4
Explanation: The subarray[1, -1, 5, -2]sums to 3 and is the longest.
Example 2:
Input: nums =[-2, -1, 2, 1], k =1
Output: 2
Explanation: The subarray[-1, 2]sums to 1 and is the longest.
Follow Up:
Can you do it in O(n) time?
题解：
用HashMap记录<sum, index> pair. 遇到sum-k在HashMap中时说明从hm.get(sum-k) 到 i 这一段的和是k.
Time Complexity: O(n). n = nums.length.
Space: O(n).
AC Java:
class Solution {
public int maxSubArrayLen(int[] nums, int k) {
if(nums == null || nums.length == 0){
return 0;
}
int sum = 0;
int res = 0;
HashMap<Integer, Integer> hm = new HashMap<Integer, Integer>();
hm.put(0, -1);
for(int i = 0; i<nums.length; i++){
sum += nums[i];
if(hm.containsKey(sum-k)){
res = Math.max(res, i-hm.get(sum-k));
}
if(!hm.containsKey(sum)){
hm.put(sum, i);
}
}
return res;
}
}