[leetcode] 915. Partition Array into Disjoint Intervals

Description

Given an array A, partition it into two (contiguous) subarrays left and right so that:

• Every element in left is less than or equal to every element in right.
• left and right are non-empty.
• left has the smallest possible size.
  Return the length of left after such a partitioning. It is guaranteed that such a partitioning exists.
  Example 1:

Input: [5,0,3,8,6]
Output: 3
Explanation: left = [5,0,3], right = [8,6]

Example 2:

Input: [1,1,1,0,6,12]
Output: 4
Explanation: left = [1,1,1,0], right = [6,12]

Note:

1. 2 <= A.length <= 30000
2. 0 <= A[i] <= 10^6
3. It is guaranteed there is at least one way to partition A as described.

分析

题目的意思是：这道题要把一个list分成两个连续的list，第一个list的所有的值都不大于第二个list，还要保证第一个list的长度最小。
我是左思右想也没想到用两个数据就搞定了，left数组就存储A[0:i]的最大值，right数据就存储A[i:]的最小值，从左到右遍历如果left[i-1]<=right[i]，说明已经找到了满足条件的i，学习一下

代码

class Solution:
    def partitionDisjoint(self, A: List[int]) -> int:
        N=len(A)
        maxleft=[None]*N
        minright=[None]*N
        
        m=A[0]
        for i in range(N):
            m=max(m,A[i])
            maxleft[i]=m
        m=A[-1]
        for i in range(N-1,-1,-1):
            m=min(m,A[i])
            minright[i]=m
        
        for i in range(1,N):
            if(maxleft[i-1]<=minright[i]):
                return i
        

参考文献

​​[LeetCode]Approach 1: Next Array​​