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.
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?
public int maxSubArrayLen(int[] nums, int k) { int sum = 0, max = 0; HashMap<Integer, Integer> map = new HashMap<Integer, Integer>(); for (int i = 0; i < nums.length; i++) { sum = sum + nums[i]; if (sum == k) max = i + 1; else if (map.containsKey(sum - k)) max = Math.max(max, i - map.get(sum - k)); if (!map.containsKey(sum)) map.put(sum, i); } return max; }reference:https://leetcode.com/discuss/77879/o-n-super-clean-9-line-java-solution-with-hashmap
转载于:https://www.cnblogs.com/hygeia/p/5123140.html
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