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這裡的Monotonic Stack不同於84的用法

stack放的是sort好的區塊，放上最大值(或是區塊的最後一個)

或是利用只有0~n-1的特性

Problem

You are given an integer array arr of length n that represents a permutation of the integers in the range [0, n - 1].

We split arr into some number of chunks (i.e., partitions), and individually sort each chunk. After concatenating them, the result should equal the sorted array.

Return the largest number of chunks we can make to sort the array.

 

Example 1:

Input: arr = [4,3,2,1,0]Output: 1Explanation: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: 4Explanation: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.

 

Constraints:

  • n == arr.length
  • 1 <= n <= 10
  • 0 <= arr[i] < n
  • All the elements of arr are unique.

Sol

class Solution:
    def maxChunksToSorted(self, arr: List[int]) -> int:
        stk = []
        for n in arr:
            maxN = n
            while stk and stk[-1] > n:
                maxN = max(maxN, stk.pop())
            stk += maxN,
        return len(stk)

但這裡可以利用只有0~n-1的特性!! 只要在對的位置上，從0~i之間，一定是最大的

class Solution:
    def maxChunksToSorted(self, arr: List[int]) -> int:
        curMax, res = -1, 0
        for i, v in enumerate(arr):
            curMax = max(curMax, v)
            res += curMax == i
        return res