[Swift]LeetCode769. 最多能完成排序的块 | Max Chunks To Make Sorted

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Given an array arr that is a permutation of [0, 1, ..., arr.length - 1], we split the array into some number of "chunks" (partitions), and individually sort each chunk.  After concatenating them, the result equals the sorted array.

What is the most number of chunks we could have made?

Example 1:

Input: arr = [4,3,2,1,0]
Output: 1
Explanation:
Splitting into two or more chunks will not return the required result.
For example, splitting into [4, 3], [2, 1, 0] will result in [3, 4, 0, 1, 2], which isn't sorted.

Example 2:

Input: arr = [1,0,2,3,4]
Output: 4
Explanation:
We can split into two chunks, such as [1, 0], [2, 3, 4].
However, splitting into [1, 0], [2], [3], [4] is the highest number of chunks possible.

Note:

• arr will have length in range [1, 10].
• arr[i] will be a permutation of [0, 1, ..., arr.length - 1].

---

数组arr是[0, 1, ..., arr.length - 1]的一种排列，我们将这个数组分割成几个“块”，并将这些块分别进行排序。之后再连接起来，使得连接的结果和按升序排序后的原数组相同。

我们最多能将数组分成多少块？

示例 1:

输入: arr = [4,3,2,1,0]
输出: 1
解释:
将数组分成2块或者更多块，都无法得到所需的结果。
例如，分成 [4, 3], [2, 1, 0] 的结果是 [3, 4, 0, 1, 2]，这不是有序的数组。

示例 2:

输入: arr = [1,0,2,3,4]
输出: 4
解释:
我们可以把它分成两块，例如 [1, 0], [2, 3, 4]。
然而，分成 [1, 0], [2], [3], [4] 可以得到最多的块数。

注意:

• arr 的长度在 [1, 10] 之间。
• arr[i]是 [0, 1, ..., arr.length - 1]的一种排列。

---

Runtime: 4 ms

Memory Usage: 19.1 MB

class Solution {
    func maxChunksToSorted(_ arr: [Int]) -> Int {
        var res:Int = 0
        var n:Int = arr.count
        var mx:Int = 0
        for i in 0..<n
        {
            mx = max(mx, arr[i])
            if mx == i
            {
                res += 1
            }
        }
        return res
    }
}

---

18268 kb

class Solution {
    func maxChunksToSorted(_ arr: [Int]) -> Int {        
        let n = arr.count
        var minOfRight = Array(repeating: 0, count: n)
        var maxOfLeft = Array(repeating: 0, count: n)
        
        maxOfLeft[0] = arr[0]
        minOfRight[n-1] = arr[n-1]
        
        for i in 1..<n {
            maxOfLeft[i] = max(maxOfLeft[i-1], arr[i])
        }
        
        for i in (0..<(n - 1)).reversed() {
            minOfRight[i] = min(minOfRight[i+1], arr[i])
        }
        
        var res = 1
        
        for i in 1..<n {
            res += maxOfLeft[i-1] <= minOfRight[i] ? 1 : 0
        }        
        return res
    }
}

---

24ms

class Solution {
    func maxChunksToSorted(_ arr: [Int]) -> Int {
        var result = 0
        let count = arr.count
        var left = 0
        var right = 0
        
        while left < count {
            right = arr[left]
            while left <= right {
                right = max(right, arr[left])
                left += 1
            }
            result += 1
        }
        return result
    }
}