325 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.

The sum of the entire nums array is guaranteed to fit within the 32-bit signed integer range.

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?

没有

没有思路，

【1，-1，5，-2，3】k = 3

prefixSum 【1， 0， 5， 3， 6】

if(map.containsKey(prefixSum - k)) {

//update the result

res = Math.max\(res, I - map.get\(prefixSum - k\)\);

}

//if it does not have this prefixSum in the map yet, save it

//problem is you have to set

//

Map.put(0,-1);

At the very beginning so that when Map get

Input:

[1,-1,5,-2,3]

3

Output:

2

Expected:

4

class Solution {
    public int maxSubArrayLen(int[] nums, int k) {
        //[-2, -1, 2, 1]
        //-2 -3 -1 0
        // k = 1
        if (nums == null || nums.length == 0) {
            return 0;
        }
        int ans = 0;
        int prefixSum = 0;
        Map<Integer, Integer> map = new HashMap<>();
        map.put(0, -1);
        for (int i = 0; i < nums.length; i++) {
            prefixSum += nums[i];
            if (map.containsKey(prefixSum - k)) {
                ans = Math.max(ans, i - map.get(prefixSum - k));
            }
            if (!map.containsKey(prefixSum)) {
                map.put(prefixSum, i);
            }
        }
        return ans;
    }
}