問題描述

915. Partition Array into Disjoint Intervals

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 

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.

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題意

給定一個數列A，將數列A從某處分開，使得左邊數列的最大值小於等於右邊數列的最小值。求滿足條件的左邊數列的最小長度。

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思路

列舉左邊數列的長度，判斷是否滿足題設條件。可以用O(n)求出max(A[0:i])和min(A[i:len-1]), i=0,1,2,…,len-1，用於判斷是否滿足條件。由於問題沒有單調性，因此只能使用線性查詢不能用二分查詢。

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程式碼

class Solution {
public:
    vector<int> lmax;
    vector<int> rmin;

    void init(vector<int> &A)
    {
        int i, len = A.size();
        lmax.clear();
        lmax.push_back(A[0]);
        for (i=1; i<len; i++)
        {
            lmax.push_back(max(lmax.back(), A[i]));
        }
        rmin.clear();
        rmin.push_back(A[len-1]);
        for (i=len-2; i>=0; i--)
        {
            rmin.push_back(min(rmin.back(), A[i]));
        }
    }
    
    int partitionDisjoint(vector<int>& A) {
        init(A);
        int i, len = A.size();
        for (i=0; i<len-1; i++)
        {
            if (lmax[i] <= rmin[len-2-i])
            {
                return i+1;
            }
        }
    }
};