[LeetCode] Continuous Subarray Sum

Given a list of non-negative numbers and a target integer k, write a function to check if the array has a continuous subarray of size at least 2 that sums up to the multiple of k,
that is, sums up to n*k where n is also an integer.

Example 1:

Input: [23, 2, 4, 6, 7],  k=6
Output: True
Explanation: Because [2, 4] is a continuous subarray of size 2 and sums up to 6.

Example 2:

Input: [23, 2, 6, 4, 7],  k=6
Output: True
Explanation: Because [23, 2, 6, 4, 7] is an continuous subarray of size 5 and sums up to 42.

Note:

The length of the array won't exceed 10,000.
You may assume the sum of all the numbers is in the range of a signed 32-bit integer.

方法一：o(n**2)

public class Solution {
public boolean checkSubarraySum(int[] nums, int k) {
if(nums.length<2) return false;
for(int i=0;i<nums.length;i++){
int sum=nums[i];
for(int j=i+1;j<nums.length;j++){
sum+=nums[j];
if(sum==0&&k==0||k!=0&&sum%k==0) return true;
}
}
return false;
}
}
方法二： 时间复杂度o(n)

public class Solution2 {
public boolean checkSubarraySum(int[] nums, int k) {
if(nums.length<2) return false;
int sum=0;
HashMap<Integer,Integer> map=new HashMap<Integer,Integer>();
map.put(0,-1);//注意
for(int i=0;i<nums.length;i++){
sum+=nums[i];
int t=k!=0?sum%k:sum;
if(map.containsKey(t)&&i-map.get(t)>1) return true;
else if(!map.containsKey(t)) map.put(t, i);
}
return false;
}
public static void main(String[] args) {
Solution2 s=new Solution2();
System.out.println(s.checkSubarraySum(new int[]{0,0}, 0));
}
}