[leetcode-915-Partition Array into Disjoint Intervals]

Posted 2021-01-11 hellowOOOrld

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.

思路：

从左到右扫描一遍数组，用一个数组记录索引i左边出现的最大值。

然后，从右到左再扫描一遍，用另外一个数组记录索引i右边出现的最小值。

最后，比较两个数组中同一个索引i处左边的最大值与右边的最小值的情况。

int partitionDisjoint(vector<int>& A)
{
    map<int,int>mpMax,mpMin;//对应的索引
    int t = A[0];
    for(int i = 0; i < A.size(); i++)
    {
        t = max(t,A[i]);
        mpMax[i] = t;
    }
    t = A[A.size()-1];
    for(int i = A.size()-1; i >= 1; i--)
    {
        t = min(t,A[i]);
        mpMin[i] = t;
    }

    for(int i = 1; i < A.size(); i++)
    {
        if(mpMax[i-1] <= mpMin[i])return i;
    }

}

 

参考：

https://leetcode.com/problems/partition-array-into-disjoint-intervals/solution/