[LeetCode 325] Maximum Size Suba

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?

Solution： 思路与#560一致，用prefixSum array 和 hashtable来做

class Solution {
    public int maxSubArrayLen(int[] nums, int k) {
        if (nums == null || nums.length == 0)
            return 0;
        
        // construct prefixSum
//         int[] prefixSum = new int[nums.length + 1];
//         int prefix = 0;
//         int index = 1;
        
//         for (int num : nums) {
//             prefix += num;
//             prefixSum[index ++] = prefix;
//         }
        
        //for each j, in prefixSum array, find the maxisum size subarray sum equals k
        // sum of subarray (i, j) = prefixSum [j] - prefixSum[i]
        Map<Integer, Integer> prefixSumVsIndex = new HashMap<> ();
        prefixSumVsIndex.put (0, -1);
            
        int maxSize = 0;
        int prefixSum = 0;
        
        for (int j = 0; j < nums.length; j++) {
            prefixSum += nums[j];
            
            if (prefixSumVsIndex.containsKey (prefixSum - k)) {
                int prefixIndex = prefixSumVsIndex.get (prefixSum - k);
                maxSize = Math.max (maxSize, j - prefixIndex);
            }
            
            // 只需记录第一个出现的prefixSum的index，因为是从左向右遍历，往后的index肯定更大，不可能是max的
            if (!prefixSumVsIndex.containsKey (prefixSum)) {
                prefixSumVsIndex.put (prefixSum, j);
            }
        }
        
        return maxSize;
    }
}