LeetCode517. 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 .

（注：每次能够选择任意多的机器，可是每次只能移动每一个机器上的一个dress，而且只能传递到它的一个相邻machine中）

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: 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   

Example2

Input: [0,3,0]

Output: 2

Explanation: 
1st move:    0 <-- 3     0    =>    1     2     0    
2nd move:    1     2 --> 0    =>    1     1     1     

Example3

Input: [0,2,0]

Output: -1

Explanation: 
It's impossible to make all the three washing machines have the same number of dresses. 

分析

首先能够根据dress的总数目来肯定最后每一个machine上的dress，若是不能整除则返回-1.

若是有解的话，咱们老是可以将一个dress从一个machine传给另外一个machine，直到每一个machine上的dress数量相同。

Since we can operate several machines at the same time, the minium number of moves is the maximum number of necessary operations on every machine.

For a single machine, necessary operations is to transfer dresses from one side to another until sum of both sides and itself reaches the average number. We can calculate (required dresses) - (contained dresses) of each side as L and R:

L > 0 && R > 0: both sides lacks dresses, and we can only export one dress from current machines at a time, so result is abs(L) + abs(R)   
L < 0 && R < 0: both sides contains too many dresses, and we can import dresses from both sides at the same time, so result is max(abs(L), abs(R))
L < 0 && R > 0 or L >0 && R < 0: the side with a larger absolute value will import/export its extra dresses from/to current machine or other side, so result is max(abs(L), abs(R))

For example, [1, 0, 5], average is 2
for 1, L = 0 * 2 - 0 = 0, R = 2 * 2 - 5= -1, result = 1　　//对于第一个元素1，由于它左边没有元素因此须要的dress是0，目前所包含的dress是0。而他右边有两台机器，因此须要的dress是2*2=4，目前所拥有的dress是0+5=5.
for 0, L = 1 * 2 - 1= 1, R = 1 * 2 - 5 = -3, result = 3
for 5, L = 2 * 2 - 1= 3, R = 0 * 2 - 0= 0, result = 3

代码

class Solution { public int findMinMoves(int[] machines) { int len = machines.length; int[] sum = new int[machines.length + 1]; for (int i = 0; i < len; i++) sum[i + 1] = sum[i] + machines[i]; if (sum[len] % len != 0) return -1; int avg = sum[len] / len; int res = 0; for (int i = 0; i < len; ++i) { int l = i * avg - sum[i]; int r = (len - i - 1) * avg - (sum[len] - sum[i] - machines[i]); if (l > 0 && r > 0) res = Math.max(res, Math.abs(l) + Math.abs(r)); else { res = Math.max(res, Math.max(l, r)); } } return res; } }

其实仍是有点不是很明白这么解的原理，也有另外一种方法http://www.javashuo.com/article/p-xnkjnkwa-bg.html，更加简单点，可是仍是没搞清除背后的数学原理是什么。