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.

题目链接：https://leetcode.com/problems/partition-array-into-disjoint-intervals/description/

Difficulty: medium

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:

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

分析

1. 定义ml为左边序列的最大值，mr为右边序列的最小值，mr初始值min(A)；
2. 循环遍历A，更新ml，mr，其中，若A[i]>mr，则不需要更新mr(这一步不做，会时间超限)，反之，mr=A[i+1:]
3. 若ml<=mr,则返回当前序号。

参考代码

class Solution:
def partitionDisjoint(self, A):
    ml=0
    index=0
    mr=min(A)
    for i in range(len(A)-1):
        ml=max(ml,A[i])
        if(A[i]==mr):
            mr=min(A[i+1:])
        if(ml<=mr):
            index=i+1
            return index