Minimum orderly adjustment

Disclaimer: This article is a blogger original article, shall not be reproduced without the bloggers allowed. https://blog.csdn.net/iov3Rain/article/details/90244077

Title Description

There are an array of integers, write a function to find out the index m and n, as long as the elements between m and n are sorted, the entire array is ordered. Note: nm should be as small as possible, that is to say, find the shortest sequence of qualifying.

Int size n of a given array A and the array, return a tuple representing the required start and end sequence. (Prosequence from zero reference position, if the original sequence of ordered to return to [0,0]). A guarantee of the elements are positive integers.

Test sample:

[1,4,6,5,9,10],6

Returns: [2,3]

 

 

Stupid way

class Rearrange {
public:
    vector<int> findSegment(vector<int> A, int n) {
        // write code here
        vector<int> res(2, 0);
        vector<int> temp(A.begin(), A.end());
        sort(A.begin(), A.end());
        for(int i = 0; i < n; ++i)
        {
            if(temp[i] != A[i])
            {
                res[0] = i;
                break;
            }
        }
        for(int i = n - 1; i >= 0; --i)
        {
            if(temp[i] != A[i])
            {
                res[1] = i;
                return res;
            }
        }
        return res;
    }
};

Traversed twice, once from left to right to find N, once from right to left to find M

(1) from left to right to find N

If the largest element preceding the current element is smaller than the current element shall be described in [the MN] and random sequence at the current element is the greatest subject to random element N so that the update superscript current element.

(2) from right to left to find M

If the current is greater than the smallest element before the element is the current element shall be described in [the MN] and random sequence at the current element is the current minimum elements so that the update target disorder current element index M

class Rearrange {
public:
    vector<int> findSegment(vector<int> A, int n) {
        // write code here
        vector<int> res(2, 0);
        int max = A[0];
        int min = A[A.size() - 1];
        for(int i = 0; i < n; ++i)
        {
            if(A[i] >= max)
            {
                max = A[i];
            }
            else
            {
                res[1] = i;
            }
        }
        for(int i = n - 1; i >= 0; --i)
        {
            if(A[i] <= min)
            {
                min = A[i];
            }
            else
            {
                res[0] = i;
            }
        }
        return res;
    }
};