# 1033. Moving Stones Until Consecutive 移动石子直到连续

@TOC

## # 题目描述

Three stones are on a number line at positions `a`, `b`, and `c`.

Each turn, you pick up a stone at an endpoint (ie., either the lowest or highest position stone), and move it to an unoccupied position between those endpoints. Formally, let's say the stones are currently at positions `x, y, z` with `x < y < z`. You pick up the stone at either position x or position z, and move that stone to an integer position `k`, with `x < k < z` and `k != y`.

The game ends when you cannot make any more moves, ie. the stones are in consecutive positions.

When the game ends, what is the minimum and maximum number of moves that you could have made? Return the answer as an length 2 array: `answer = [minimum_moves, maximum_moves]`

Example 1:

``````Input: a = 1, b = 2, c = 5
Output: [1,2]
Explanation: Move the stone from 5 to 3, or move the stone from 5 to 4 to 3.
``````

Example 2:

``````Input: a = 4, b = 3, c = 2
Output: [0,0]
Explanation: We cannot make any moves.
``````

Example 3:

``````Input: a = 3, b = 5, c = 1
Output: [1,2]
Explanation: Move the stone from 1 to 4; or move the stone from 1 to 2 to 4.
``````

Note:

1. `1 <= a <= 100`
2. `1 <= b <= 100`
3. `1 <= c <= 100`
4. `a != b, b != c, c != a`

## # 题目大意

a,b,c表示三个位置，在三个位置上各有一个石头。现在要移动三个石头中的若干个，每次移动都必须选两端石头的里面的位置，最终使得它们三个放在连续的位置。问最少需要多少次移动，最多需要多少次移动。

## # 解题方法

### # 脑筋急转弯

C++代码如下：

``````class Solution {
public:
vector<int> numMovesStones(int a, int b, int c) {
int sum_ = a + b + c;
int min_ = min(a, min(b, c));
int max_ = max(a, max(b, c));
int mid_ = sum_ - min_ - max_;

if (max_ - min_ == 2)
return {0, 0};

int min_move = min(mid_ - min_, max_ - mid_) <= 2 ? 1 : 2;
int max_move = max_ - min_ - 2;
return {min_move, max_move};
}
};
``````

## # 日期

2019 年 8 月 31 日 —— 赶在月底做个题