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.

思路：

因为是非负数，遍历数组一遍一直加，并且记录mod k的结果，只要出现过，判断一下位置即可返回结果。

class Solution {
public boolean checkSubarraySum(int[] nums, int k) {
Map<Integer, Integer> map = new HashMap<Integer, Integer>(){{
put(0,-1);
}};
int runningSum = 0;
for (int i = 0; i < nums.length; i++) {
runningSum += nums[i];
if (k != 0)
runningSum %= k;
Integer prev = map.get(runningSum);
if (prev != null) {
if (i - prev > 1)
return true;
} else
map.put(runningSum, i);
}
return false;
}
}