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.
题目
给定数组和一个数K，求是否存在子数组和为K的倍数。
思路
代码
class Solution {
public boolean checkSubarraySum(int[] nums, int k) {
HashMap<Integer, Integer> map = new HashMap<>();
//为何 map.put(0, -1) 呢？ 如果在第2位找到了mod == 0的数，那就 1 -（-1）>1，return true。
map.put(0, -1);
int sum = 0;
for (int i = 0; i < nums.length; i++) {
// running sum
sum += nums[i];
if (k != 0) {
sum %= k;
}
// find sum % k is in the HashMap
if (map.containsKey(sum)) {
// subarray length at least two
if (i - map.get(sum) > 1) {
return true;
}
}
else {
// key: runnng sum -- value: index
map.put(sum, i);
}
}
return false;
}
}