[LeetCode]Max Chunks To Make Sorted (ver. 1)

题目描述:

LeetCode 769. Max Chunks To Make Sorted (ver. 1)

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].

题目大意:

将数组arr拆分成若干部分,使得每一部分分别排序后,重新组合在一起的数组递增有序。

求最多可以分成多少个部分。

解题思路:

贪心算法(Greedy Algorithm)

记数组arr长度为N,下标x从N - 1到0逐一递减:

    若min(arr[x .. N - 1]) > max(arr[0 .. x - 1]),则令结果+1

Python代码:

class Solution(object):
    def maxChunksToSorted(self, arr):
        """
        :type arr: List[int]
        :rtype: int
        """
        N = len(arr)
        ans = 1
        for x in range(N - 1, 0, -1):
            if min(arr[x:]) > max(arr[:x]):
                ans += 1
        return ans

 

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