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.
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()-,mina=A[A.size()-],pos=A.size()-;
while(pa>=)
{
if(A[pa]>A[pa+])
{
pos=pa;
break;
}
pa--;
}
if(pos==A.size()-)
return A;
int pos_=pos+,maxa=-,max_pos=-;
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;
}
};