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.
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 =3Output: 4 Explanation: The subarray[1, -1, 5, -2]sums to 3 and is the longest.
Example 2:
Input: nums =[-2, -1, 2, 1], k =1Output: 2 Explanation: The subarray[-1, 2]sums to 1 and is the longest.
Follow Up:
Can you do it in O(n) time?
---------------------Can you do it in O(n) time?
Solution #1
Map里key = [0 - i] 的和,value = i.
因为要找的是最左边的点(与当前i距离最远),所以遇到重复的key可以跳过
class Solution {
public int maxSubArrayLen(int[] nums, int k) {
Map<Integer, Integer> map = new HashMap<>();
int len = 0;
int sofar = 0;
for (int i = 0; i < nums.length; i++) {
sofar += nums[i];
if (sofar == k) len = i + 1;
if (map.containsKey(sofar - k)) {
len = Math.max(len, i - map.get(sofar - k));
}
if (!map.containsKey(sofar)) map.put(sofar, i);
}
return len;
}
}
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