leetcode_1053. Previous Permutation With One Swap

1053. Previous Permutation With One Swap

https://leetcode.com/problems/previous-permutation-with-one-swap/

题意：Given an array A of positive integers (not necessarily distinct), return the lexicographically largest permutation that is smaller than A, that can be made with one swap (A swap exchanges the positions of two numbers A[i] and A[j]).  If it cannot be done, then return the same array.

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解法：对于每个A，找到最右边的第一个逆序对{ A[i]>A[j] }，然后将A[i]和其后小于A[i]的最大的元素交换。

 

class Solution
{
public:
    vector<int> prevPermOpt1(vector<int>& A)
    {
        int pa=A.size()-2,mina=A[A.size()-1],pos=A.size()-1;
        while(pa>=0)
        {
            if(A[pa]>A[pa+1])
            {
                pos=pa;
                break;
            }
            pa--;
        }
        if(pos==A.size()-1)
            return A;
        int pos_=pos+1,maxa=-1,max_pos=-1;
        while(pos_<A.size())
        {
            if(A[pos_]<A[pos]&&A[pos_]>maxa)
            {
                maxa=A[pos_];
                max_pos=pos_;
            }
            pos_++;
        }
        swap(A[pos],A[max_pos]);
        return A;
    }
};