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:
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?
一刷
題解：我們把所有從0到i的sum存到map中萎津，map.put(sum, i+1), 并且如果有重復的sum, 不更新校仑，保證map中的value最小。
這樣的話颜懊，如果當前的sum!=k, 但是sum-k在map中溺忧，表示有一段subarray可以等于k, 于是max = Math.max(max, i+1-map.get(sum-k));, 這就是map不更新微渠，保證map保存最短的subarray
class Solution {
public int maxSubArrayLen(int[] nums, int k) {
int sum = 0, max = 0;
Map<Integer, Integer> map = new HashMap<>();
for(int i=0; i<nums.length; i++){
sum += nums[i];
if(sum == k) max = i+1;
else if(map.containsKey(sum - k)) max = Math.max(max, i+1-map.get(sum-k));
if(!map.containsKey(sum)) map.put(sum, i+1);
}
return max;
}
}
