523. 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.

这里用hashmap存储之前所有的子数组和subsum跟k的余数。subsum+=nums[i]，每次再跟k取余。然后在map里找有没有这个余数，有的话再检查是不是符合至少间隔两个index，还符合的话返回true，不符合的话，存储当前余数和index。还有一点要注意，取余运算要检查除数是否为0。代码如下：
public class Solution {
public boolean checkSubarraySum(int[] nums, int k) {
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
map.put(0, -1);
int sum = 0;
for (int i = 0; i < nums.length; i ++) {
sum += nums[i];
if (k != 0)
sum %= k;
if (map.containsKey(sum)) {
if (i - map.get(sum) > 1) {
return true;
}
} else {
map.put(sum, i);
}
}
return false;
}
}