517. Super Washing Machines

517. Super Washing Machines

You have n super washing machines on a line. Initially, each washing machine has some dresses or is empty.

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 .

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.

Example1

Input: [1,0,5]Output: 3Explanation: 1st move: 1 0 <-- 5 => 1 1 42nd move: 1 <-- 1 <-- 4 => 2 1 3 3rd move: 2 1 <-- 3 => 2 2 2

Example2

Input: [0,3,0]Output: 2Explanation: 1st move: 0 <-- 3 0 => 1 2 0 2nd move: 1 2 --> 0 => 1 1 1

Example3

Input: [0,2,0]Output: -1Explanation: It's impossible to make all the three washing machines have the same number of

Note: The range of n is [1, 10000]. The range of dresses number in a super washing machine is [0, 1e5].

思路: 根据本题题意，首先，要达到目标是使得每台洗衣机具有一样的衣服数，这意味着所有洗衣机的衣服总数是所有洗衣机个数的倍数，那么sum%size应该为0，这构成了第一个条件，而且达到平衡时的衣服数为target=sum/size。另外，每台洗衣机每次只能移动一件衣服到邻接的洗衣机处，那么最小的移动次数将会与最多衣服的洗衣机有关。当只有一台洗衣机的衣服最多时，可以很容易得出minmove=max-target，当拥有最多衣服的洗衣机不止一台时，此时需要找出达到平衡时需要移动最多步数的洗衣机，而计算在每台洗衣机移动步数时，可引入balance变量来记录达到平衡需要的步数，使用累加的方法，可以巧妙地计算到每台洗衣机的移动步数，从左开始算，负数代表从左移动的步数，正数则是从右移动的步数，那么其绝对值则是最少移动步数。上述两种方法结合，再求最大值，就可得出真正的最少步数。

class Solution { public int findMinMoves(int[] machines) { int sum = 0, length = machines.length; for(int i=0; i

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