题目描述：
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.
思路：
思路：暴力解之。注意 k 为 0 时应单独考虑，判断相邻的元素是否为0，若是返回 true ，否则返回 false 。
注：此题有 O(n) 的解法，借助数学特性，具体见解法 2 。
代码实现：
解法 1 ：暴力
class Solution {
public boolean checkSubarraySum(int[] nums, int k) {
if(k==0){
return checkzero(nums);
}
int len = nums.length;
for(int i=0;i<len-1;i++){
int sum=0;
for(int j=i;j<len;j++){
sum+=nums[j];
if(sum%k==0 && j-i>0){
return true;
}
}
}
return false;
}
private boolean checkzero(int[] nums){
for(int i=0;i<nums.length-1;i++){
if(nums[i]==0 && nums[i+1]==0){
return true;
}
}
return false;
}
}
解法 2 ：时间复杂度 O(n)
class Solution {
public:
bool checkSubarraySum(vector<int>& nums, int k) {
int n = nums.size(), sum = 0, pre = 0;
unordered_set<int> modk;
for (int i = 0; i < n; ++i) {
sum += nums[i];
int mod = k == 0 ? sum : sum % k;
if (modk.count(mod)) return true;
modk.insert(pre);
pre = mod;
}
return false;
}
};