[Swift Weekly Contest 118]LeetCode974. 和可被 K 整除的子數組 | Subarray Sums Divisible by K
阿新 • • 發佈:2019-01-13
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Given an array A
of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by K
.
Example 1:
Input: A = [4,5,0,-2,-3,1], K = 5
Output: 7
Explanation: There are 7 subarrays with a sum divisible by K = 5:
[4, 5, 0, -2, -3, 1], [5], [5, 0], [5, 0, -2, -3], [0], [0, -2, -3], [-2, -3]
Note:
1 <= A.length <= 30000
-10000 <= A[i] <= 10000
2 <= K <= 10000
給定一個整數數組 A
,返回其中元素之和可被 K
整除的(連續、非空)子數組的數目。
示例:
輸入:A = [4,5,0,-2,-3,1], K = 5 輸出:7 解釋: 有 7 個子數組滿足其元素之和可被 K = 5 整除: [4, 5, 0, -2, -3, 1], [5], [5, 0], [5, 0, -2, -3], [0], [0, -2, -3], [-2, -3]
提示:
1 <= A.length <= 30000
-10000 <= A[i] <= 10000
2 <= K <= 10000
368ms
1 class Solution { 2 func subarraysDivByK(_ A: [Int], _ K: Int) -> Int { 3 var n = A.count 4 var f:[Int] = [Int](repeating:0,count:K) 5 f[0] = 1 6 var x:Int = 0 7 for v in A 8{ 9 x += v 10 x %= K 11 if x < 0 12 { 13 x += K 14 } 15 f[x] += 1 16 } 17 var ret:Int = 0 18 for i in 0..<K 19 { 20 ret += f[i]*(f[i]-1)/2 21 } 22 return ret 23 } 24 }
[Swift Weekly Contest 118]LeetCode974. 和可被 K 整除的子數組 | Subarray Sums Divisible by K