523. Continuous Subarray Sum ⭐

Medium
​
Given an integer array nums and an integer k, return true if nums has a continuous subarray of size at least two whose elements sum up to a multiple of k, or false otherwise.
An integer x is a multiple of k if there exists an integer n such that x = n * k. 0 is always a multiple of k.
Example 1:
Input: nums = [23,2,4,6,7], k = 6
Output:
true
Explanation:
[2, 4] is a continuous subarray of size 2 whose elements sum up to 6.
Example 2:
Input: nums = [23,2,6,4,7], k = 6
Output:
true
Explanation:
[23, 2, 6, 4, 7] is an continuous subarray of size 5 whose elements sum up to 42.
42 is a multiple of 6 because 42 = 7 * 6 and 7 is an integer.
Example 3:
Input: nums = [23,2,6,4,7], k = 13
Output:
false
Constraints:
  • 1 <= nums.length <= 105
  • 0 <= nums[i] <= 109
  • 0 <= sum(nums[i]) <= 231 - 1
  • 1 <= k <= 231 - 1
​

解題

這一題使用到前綴和的概念與mod的概念。首先,我們建立一個map來記錄目前 sum % k 的餘數,如果該數出現第二遍,代表 上一次出現該餘數的 index+1 到目前index 的值加總等於 k ,所以餘數才會再次出現。
func checkSubarraySum(nums []int, k int) bool {
m := make(map[int]int)
m[0] = -1
sum := 0
for i, val := range nums {
sum += val
if k != 0 { sum %= k }
if preindex, ok := m[sum]; ok {
if i-preindex>1 { return true }
} else {
m[sum] = i
}
}
return false
}