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➤微信公衆號:山青詠芝(shanqingyongzhi)
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Given an array A
of positive integers, let S
be the sum of the digits of the minimal element of A
.git
Return 0 if S
is odd, otherwise return 1.github
Example 1:數組
Input: [34,23,1,24,75,33,54,8]
Output: 0 Explanation: The minimal element is 1, and the sum of those digits is S = 1 which is odd, so the answer is 0.
Example 2:微信
Input: [99,77,33,66,55]
Output: 1 Explanation: The minimal element is 33, and the sum of those digits is S = 3 + 3 = 6 which is even, so the answer is 1.
Note:測試
1 <= A.length <= 100
1 <= A[i].length <= 100
給你一個正整數的數組 A
。spa
而後計算 S
,使其等於數組 A
當中最小的那個元素各個數位上數字之和。code
最後,假如 S
所得計算結果是 奇數 的請你返回 0,不然請返回 1。htm
示例 1:blog
輸入:[34,23,1,24,75,33,54,8] 輸出:0 解釋: 最小元素爲 1,該元素各個數位上的數字之和 S = 1,是奇數因此答案爲 0。
示例 2:
輸入:[99,77,33,66,55] 輸出:1 解釋: 最小元素爲 33,該元素各個數位上的數字之和 S = 3 + 3 = 6,是偶數因此答案爲 1。
提示:
1 <= A.length <= 100
1 <= A[i].length <= 100
32ms
1 class Solution { 2 func sumOfDigits(_ A: [Int]) -> Int { 3 var m:Int = Int.max 4 for p in A 5 { 6 m = min(p,m) 7 } 8 var s:Int = 0 9 while(m != 0) 10 { 11 s += m%10 12 m /= 10 13 } 14 return (s%2)^1 15 } 16 }