[Swift]LeetCode1085. 最小元素各數位之和 | Sum of Digits in the Minimum Number

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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. 1 <= A.length <= 100
  2. 1 <= A[i].length <= 100

給你一個正整數的數組 Aspa

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. 1 <= A.length <= 100
  2. 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 }
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