[Swift]LeetCode769. 最多能完成排序的塊 | 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].

---

數組arr是[0, 1, ..., arr.length - 1]的一種排列，咱們將這個數組分割成幾個「塊」，並將這些塊分別進行排序。以後再鏈接起來，使得鏈接的結果和按升序排序後的原數組相同。

咱們最多能將數組分紅多少塊？

示例 1:

輸入: arr = [4,3,2,1,0]
輸出: 1
解釋:
將數組分紅2塊或者更多塊，都沒法獲得所需的結果。
例如，分紅 [4, 3], [2, 1, 0] 的結果是 [3, 4, 0, 1, 2]，這不是有序的數組。

示例 2:

輸入: arr = [1,0,2,3,4]
輸出: 4
解釋:
咱們能夠把它分紅兩塊，例如 [1, 0], [2, 3, 4]。
然而，分紅 [1, 0], [2], [3], [4] 能夠獲得最多的塊數。

注意:

• arr 的長度在 [1, 10] 之間。
• arr[i]是 [0, 1, ..., arr.length - 1]的一種排列。

---

Runtime: 4 ms

Memory Usage: 19.1 MB

 1 class Solution {
 2     func maxChunksToSorted(_ arr: [Int]) -> Int {
 3         var res:Int = 0
 4         var n:Int = arr.count
 5         var mx:Int = 0
 6         for i in 0..<n
 7         {
 8             mx = max(mx, arr[i])
 9             if mx == i
10             {
11                 res += 1
12             }
13         }
14         return res
15     }
16 }

---

18268 kb

 1 class Solution {
 2     func maxChunksToSorted(_ arr: [Int]) -> Int {        
 3         let n = arr.count
 4         var minOfRight = Array(repeating: 0, count: n)
 5         var maxOfLeft = Array(repeating: 0, count: n)
 6         
 7         maxOfLeft[0] = arr[0]
 8         minOfRight[n-1] = arr[n-1]
 9         
10         for i in 1..<n {
11             maxOfLeft[i] = max(maxOfLeft[i-1], arr[i])
12         }
13         
14         for i in (0..<(n - 1)).reversed() {
15             minOfRight[i] = min(minOfRight[i+1], arr[i])
16         }
17         
18         var res = 1
19         
20         for i in 1..<n {
21             res += maxOfLeft[i-1] <= minOfRight[i] ? 1 : 0
22         }        
23         return res
24     }
25 }

---

24ms

 1 class Solution {
 2     func maxChunksToSorted(_ arr: [Int]) -> Int {
 3         var result = 0
 4         let count = arr.count
 5         var left = 0
 6         var right = 0
 7         
 8         while left < count {
 9             right = arr[left]
10             while left <= right {
11                 right = max(right, arr[left])
12                 left += 1
13             }
14             result += 1
15         }
16         return result
17     }
18 }