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?
代码
Approach #1 HashMap
class Solution {
public int maxSubArrayLen(int[] nums, int k) {
int sum = 0, max = 0;
HashMap<Integer, Integer> map = new HashMap();
for (int i = 0; i < nums.ength; 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;
}
}