题目:
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:
我们定义一个哈希表,用来保存截止目前的和模k之后的余数,例如截止i的时候,哈希表里面有2,3两个数,那么就表示扫描到nums[i]的时候,前面的和模k之后的余数有2,3。那么当我们扫描到j >= i + 2或者之后,如果发现截止当前的和模k的余数已经在哈希表里面了,就说明从[i + 1, j]这个闭区间的数的和可以被k整除。
一般情况下,我们可以用一个unordered_map,顺便把索引也保存一下,这样便于判断两个索引的差值是不是大于等于2。但是在下面的代码片段里面我们用了一个过渡变量pre,也就是扫描到当前位置的时候,我们才插入前一个余数pre,这样其实只需要一个unordered_set就可以了。
算法的时间复杂度是O(n),空间复杂度是O(k),因为hash表的大小的最大可能值就是k。
代码:
class Solution {
public:
bool checkSubarraySum(vector<int>& nums, int k) {
int sum = 0, pre = 0;
unordered_set<int> hash;
for (int i = 0; i < nums.size(); ++i) {
sum += nums[i];
sum = (k == 0) ? sum : sum % k;
if (hash.count(sum)) {
return true;
}
hash.insert(pre);
pre = sum;
}
return false;
}
};