LeetCode[517] Super Washing Machines

最新推荐文章于 2021-09-30 10:30:06 发布

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本文介绍了一种通过最少步骤使每台超级洗衣机内的衣物数量相等的算法。利用动态规划思想，该算法首先计算所有洗衣机中衣物总数，并检查是否能平均分配。接着遍历每台洗衣机，累计偏差并更新总移动次数。

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;
    }
};