LeetCode[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: 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.

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

此题不难理解，简单分析的话，动态规划是个不错的解法。

class Solution {public:int findMinMoves(vector<int>& machines) {    int totalDresses = 0, size = machines.size();    for (auto i = 0; i < size; ++i) totalDresses += machines[i];    if (totalDresses % size != 0) return -1;    auto targetDresses = totalDresses / size, totalMoves = 0, ballance = 0;    for (auto i = 0; i < size; ++i) {        ballance += machines[i] - targetDresses;        totalMoves = max(totalMoves, max(machines[i] - targetDresses, abs(ballance)));    }    return totalMoves;    }};

上面的代码呢是我按别人的修改后的结果。不难看出还是很简洁的，个人感觉有了auto，能省不少时间，可以少花些时间在变量类型的判断上。