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
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.
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.
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.
Note
If presum1%k = c and presum2%k = c then |presum1 - presum2|%k=0
注意一是0的位置在这里得初始化为-1,二是注意0的情况[0, 0], k = 0,输出是true,如果k是0的话就不取模了,三是注意间隔大于1,意味着presume的差不是单个元素。
Math.floorMod(sum, k) ---> 不会给出负数结果
Code
class Solution {
public boolean checkSubarraySum(int[] nums, int k) {
int sum = 0;
Map<Integer, Integer> map = new HashMap<>();
map.put(0, -1);
for (int i = 0; i < nums.length; i++) {
sum += nums[i];
int remainder = sum;
if (k != 0) {
remainder = sum % k;
}
if (map.containsKey(remainder)) {
if (i - map.get(remainder) > 1) {
return true;
}
} else {
map.put(remainder, i);
}
}
return false;
}
}
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