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LeetCode

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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.

Follow Up:

Can you do it in O(n) time?

Code

把当前和以及最小index保留在map中，然后用k - nums[i]找map中的值

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;
    }
}