[Swift]LeetCode1085. 最小元素各数位之和 | 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
.
Return 0 if S
is odd, otherwise return 1.
Example 1:
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:
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:
1 <= A.length <= 100
1 <= A[i].length <= 100
给你一个正整数的数组 A
。
然后计算 S
,使其等于数组 A
当中最小的那个元素各个数位上数字之和。
最后,假如 S
所得计算结果是 奇数 的请你返回 0,否则请返回 1。
示例 1:
输入:[34,23,1,24,75,33,54,8] 输出:0 解释: 最小元素为 1,该元素各个数位上的数字之和 S = 1,是奇数所以答案为 0。
示例 2:
输入:[99,77,33,66,55] 输出:1 解释: 最小元素为 33,该元素各个数位上的数字之和 S = 3 + 3 = 6,是偶数所以答案为 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 }