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You have n super washing machines on a line. Initially, each washing machine has some dresses or is empty.git
For each move, you could choose any m (1 ≤ m ≤ n) washing machines, and pass one dress of each washing machine to one of its adjacent washing machines at the same time .github
Given an integer array representing the number of dresses in each washing machine from left to right on the line, you should find the minimum number of moves to make all the washing machines have the same number of dresses. If it is not possible to do it, return -1.数组
Example1微信
Input: [1,0,5] Output: 3 Explanation: 1st move: 1 0 <-- 5 => 1 1 4 2nd move: 1 <-- 1 <-- 4 => 2 1 3 3rd move: 2 1 <-- 3 => 2 2 2
Example2spa
Input: [0,3,0] Output: 2 Explanation: 1st move: 0 <-- 3 0 => 1 2 0 2nd move: 1 2 --> 0 => 1 1 1
Example3code
Input: [0,2,0] Output: -1 Explanation: It's impossible to make all the three washing machines have the same number of dresses.
Note:htm
假设有 n 台超级洗衣机放在同一排上。开始的时候,每台洗衣机内可能有必定量的衣服,也多是空的。blog
在每一步操做中,你能够选择任意 m (1 ≤ m ≤ n) 台洗衣机,与此同时将每台洗衣机的一件衣服送到相邻的一台洗衣机。three
给定一个非负整数数组表明从左至右每台洗衣机中的衣物数量,请给出能让全部洗衣机中剩下的衣物的数量相等的最少的操做步数。若是不能使每台洗衣机中衣物的数量相等,则返回 -1。
示例 1:
输入: [1,0,5] 输出: 3 解释: 第一步: 1 0 <-- 5 => 1 1 4 第二步: 1 <-- 1 <-- 4 => 2 1 3 第三步: 2 1 <-- 3 => 2 2 2
示例 2:
输入: [0,3,0] 输出: 2 解释: 第一步: 0 <-- 3 0 => 1 2 0 第二步: 1 2 --> 0 => 1 1 1
示例 3:
输入: [0,2,0] 输出: -1 解释: 不可能让全部三个洗衣机同时剩下相同数量的衣物。
提示:
1 class Solution { 2 func findMinMoves(_ machines: [Int]) -> Int { 3 var sum:Int = machines.reduce(0, +) 4 if sum % machines.count != 0 {return -1} 5 var res:Int = 0 6 var cnt:Int = 0 7 var avg:Int = sum / machines.count 8 for m in machines 9 { 10 cnt += m - avg 11 res = max(res, max(abs(cnt), m - avg)) 12 } 13 return res 14 } 15 }
1 class Solution { 2 func findMinMoves(_ machines: [Int]) -> Int { 3 var total = 0 4 for machine in machines { 5 total += machine 6 } 7 8 if (total % machines.count != 0) { 9 return -1 10 } else { 11 let average = total / machines.count 12 var dis = 0 13 var result = 0 14 15 for machine in machines { 16 let currentDis = machine - average 17 dis += currentDis 18 result = max(result, abs(dis)) 19 result = max(result, currentDis) 20 } 21 return result 22 } 23 } 24 }