LeetCode每日一题(517. Super Washing Machines)-优快云博客

给定一个整数数组 machines，表示从左到右每台洗衣机上的衣服数量。目标是通过最少的操作次数，使得所有洗衣机上的衣服数量相等。每次操作可以选择任意 m (1 <= m <= n) 台洗衣机，并同时向相邻的一台洗衣机传递一件衣服。如果无法使所有洗衣机衣服数量相同，则返回 -1。例如，输入 [1,0,5]，输出为 3，经过三次操作，所有洗衣机都有两件衣服。" 117109109,10293547,奇数阶幻方构造：C语言实现与算法解析,"['算法', 'C语言', '数学', '数值计算']

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 machines representing the number of dresses in each washing machine from left to right on the line, return 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.

Example 1:

Input: machines = [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

Example 2:

Input: machines = [0,3,0]
Output: 2

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

Example 3:

Input: machines = [0,2,0]
Output: -1

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

Constraints:

n == machines.length
1 <= n <= 104
0 <= machines[i] <= 105

看了人家发上去的解析, 想了半天，感觉有点理解了。其实我们要找的是一个向左右两边都输出的机器，当然这个左右两边实际上包括左边为空或者右边为空的情况。而且这个最大值一定是存在的。不好描述，大家还是看原文自己理解吧。


impl Solution {
    pub fn find_min_moves(machines: Vec<i32>) -> i32 {
        let mut sum = 0;
        let length = machines.len();
        let mut prefix_sum = vec![0; machines.len() + 1];
        for (i, m) in machines.into_iter().enumerate() {
            sum += m;
            prefix_sum[i + 1] = prefix_sum[i] + m;
        }
        if sum % length as i32 != 0 {
            return -1;
        }
        let avg = sum / length as i32;
        let mut ans = 0;
        for i in 0..length {
            let left_diff = i as i32 * avg - (prefix_sum[i] - prefix_sum[0]);
            let right_diff =
                (length - i - 1) as i32 * avg - (prefix_sum[length] - prefix_sum[i + 1]);
            let left = left_diff.clamp(0, i32::MAX);
            let right = right_diff.clamp(0, i32::MAX);
            ans = ans.max(left + right);
        }
        ans
    }
}