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?
首先subarray要连续,这题就跟之前range sum差不多。
需要用map记录当前位置之前的和。
code:
public class Solution {
public int maxSubArrayLen(int[] nums, int k) {
Map<Integer,Integer> mp = new HashMap<Integer, Integer>();
if(nums == null || nums.length == 0) return 0;
int len = 0, sum = 0;
for(int i = 0; i<nums.length; ++i){
sum += nums[i];
if(sum == k) len = i+1;
if(mp.containsKey(sum-k)){
len = Math.max(len, i- mp.get(sum-k));
}
if(mp.containsKey(sum)) mp.put(sum, mp.get(sum));// if there already have sum, we should keep its idx, for longest array reason: [1,0,-1] k = -1, return 2;
else mp.put(sum, i);
}
return len;
}
}