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On an infinite number line, the position of the i-th stone is given by stones[i]. Call a stone an endpoint stone if it has the smallest or largest position.git
Each turn, you pick up an endpoint stone and move it to an unoccupied position so that it is no longer an endpoint stone.github
In particular, if the stones are at say, stones = [1,2,5], you cannotmove the endpoint stone at position 5, since moving it to any position (such as 0, or 3) will still keep that stone as an endpoint stone.数组
The game ends when you cannot make any more moves, ie. the stones are in consecutive positions.微信
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]app
Example 1:spa
Input: [7,4,9]
Output: [1,2] Explanation: We can move 4 -> 8 for one move to finish the game. Or, we can move 9 -> 5, 4 -> 6 for two moves to finish the game.
Example 2:code
Input: [6,5,4,3,10]
Output: [2,3] We can move 3 -> 8 then 10 -> 7 to finish the game. Or, we can move 3 -> 7, 4 -> 8, 5 -> 9 to finish the game. Notice we cannot move 10 -> 2 to finish the game, because that would be an illegal move.
Example 3:htm
Input: [100,101,104,102,103]
Output: [0,0]
Note:blog
3 <= stones.length <= 10^41 <= stones[i] <= 10^9stones[i] have distinct values.在一个长度无限的数轴上，第 i 颗石子的位置为 stones[i]。若是一颗石子的位置最小/最大，那么该石子被称做端点石子。
每一个回合，你能够将一颗端点石子拿起并移动到一个未占用的位置，使得该石子再也不是一颗端点石子。
值得注意的是，若是石子像 stones = [1,2,5] 这样，你将没法移动位于位置 5 的端点石子，由于不管将它移动到任何位置（例如 0 或 3），该石子都仍然会是端点石子。
当你没法进行任何移动时，即，这些石子的位置连续时，游戏结束。
要使游戏结束，你能够执行的最小和最大移动次数分别是多少？ 以长度为 2 的数组形式返回答案：answer = [minimum_moves, maximum_moves] 。
示例 1：
输入：[7,4,9] 输出：[1,2] 解释： 咱们能够移动一次，4 -> 8，游戏结束。 或者，咱们能够移动两次 9 -> 5，4 -> 6，游戏结束。
示例 2：
输入：[6,5,4,3,10] 输出：[2,3] 解释： 咱们能够移动 3 -> 8，接着是 10 -> 7，游戏结束。 或者，咱们能够移动 3 -> 7, 4 -> 8, 5 -> 9，游戏结束。 注意，咱们没法进行 10 -> 2 这样的移动来结束游戏，由于这是不合要求的移动。
示例 3：
输入：[100,101,104,102,103] 输出：[0,0]
提示：
3 <= stones.length <= 10^41 <= stones[i] <= 10^9stones[i] 的值各不相同。1 class Solution { 2 func numMovesStonesII(_ stones: [Int]) -> [Int] { 3 let n = stones.count, sortStones = stones.sorted() 4 var maxcontinuous = 1, end = 0, j = 0 5 var si = 0, ei = 0 6 for i in 0..<n { 7 end = sortStones[i] + n 8 while j < n && sortStones[j] < end { j += 1 } 9 10 if j - i > maxcontinuous || (j - i == maxcontinuous && sortStones[j-1] - sortStones[i] > ei - si ) { 11 si = sortStones[i] 12 ei = sortStones[j-1] 13 maxcontinuous = j-i 14 } 15 } 16 17 var minimum = n - maxcontinuous 18 if maxcontinuous == n-1 && ei-si == maxcontinuous-1 { minimum += 1 } 19 if maxcontinuous == n { minimum = 0 } 20 let largest_interval = max(sortStones[n-1] - sortStones[1], sortStones[n-2] - sortStones[0]) - 1 21 let maximum = largest_interval - ( n - 3) 22 return [minimum, maximum] 23 } 24 }
160ms
1 class Solution { 2 private var stones: [Int] = [] 3 4 func numMovesStonesII(_ stones: [Int]) -> [Int] { 5 self.stones = stones.sorted() 6 return [findMin(), findMax()] 7 } 8 9 private func findMax() -> Int { 10 var count = 0 11 var last: Int = stones[0] 12 13 for i in 1..<(stones.count - 1) { 14 count += stones[i] - last - 1 15 last = stones[i] 16 } 17 18 var result = count 19 20 count = 0 21 last = stones[1] 22 for i in 2..<stones.count { 23 count += stones[i] - last - 1 24 last = stones[i] 25 } 26 27 result = max(result, count) 28 29 return result 30 } 31 32 private func findMin() -> Int { 33 var window = (left: 0, right: 1) 34 var minMoves = Int.max 35 36 while true { 37 let windowSpan = stones[window.right] - stones[window.left] + 1 38 let windowSize = window.right - window.left + 1 39 if windowSpan <= stones.count { 40 if windowSpan == windowSize && windowSize == stones.count - 1 { 41 if stones[stones.count - 1] - stones[0] == stones.count - 1 { 42 return 0 43 } else { 44 if window.left == 1 { 45 return stones[1] - stones[0] == 2 ? 1 : 2 46 } else { 47 return stones[stones.count - 1] - stones[window.right] == 2 ? 1 : 2 48 } 49 } 50 } 51 52 minMoves = min(minMoves, stones.count - (window.right - window.left + 1)) 53 if window.right != stones.count - 1 { 54 window.right += 1 55 } else { 56 break 57 } 58 } else { 59 window.left += 1 60 } 61 } 62 63 return minMoves 64 } 65 }
Runtime: 192 ms
1 class Solution { 2 func numMovesStonesII(_ stones: [Int]) -> [Int] { 3 var stones = stones.sorted(by:<) 4 var mx:Int = 0 5 var n:Int = stones.count 6 for i in 1..<n 7 { 8 mx += stones[i] - stones[i - 1] - 1 9 } 10 if mx == 0 {return [0,0]} 11 mx -= min(stones[1] - stones[0] - 1, stones[n - 1] - stones[n - 2] - 1) 12 var res:Int = 0 13 if (stones[n - 2] - stones[0] == n - 2 && stones[n - 1] - stones[n - 2] > 2) || (stones[n - 1] - stones[1] == n - 2 && stones[1] - stones[0] > 2) 14 { 15 return [2, mx] 16 } 17 else 18 { 19 var dq:[Int] = [Int]() 20 for stone in stones 21 { 22 dq.append(stone) 23 while(dq.last! - dq.first! + 1 > n) 24 { 25 dq.removeFirst() 26 } 27 res = max(res, dq.count) 28 } 29 } 30 return [n - res, mx] 31 } 32 }