• ㊗️
• 大家
• offer
• 多多！

## Problem

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.


Can you do it in O(n) time?

## Code

class Solution {
public int maxSubArrayLen(int[] nums, int k) {
if (nums == null || nums.length == 0) return 0;

HashMap<Integer, Integer> map = new HashMap<>();
map.put(0, -1);

// 保证0的index是-1
if (nums[0] != 0){
map.put(nums[0], 0);
}

for (int i = 1; i < nums.length; i++) {
nums[i] += nums[i - 1];

if (!map.containsKey(nums[i])){
map.put(nums[i], i);
}
}

int res = 0;

for(int i = 0; i < nums.length; i++) {
if(map.containsKey(nums[i] - k)){
res = Math.max(res, i - map.get(nums[i] - k));
}
}

return res;
}
}