title
We have some permutation A of [0, 1, …, N - 1], where N is the length of A.
The number of (global) inversions is the number of i < j with 0 <= i < j < N and A[i] > A[j].
The number of local inversions is the number of i with 0 <= i < N and A[i] > A[i+1].
Return true if and only if the number of global inversions is equal to the number of local inversions.
Example 1:
Input: A = [1,0,2]
Output: true
Explanation: There is 1 global inversion, and 1 local inversion.
Example 2:
Input: A = [1,2,0]
Output: false
Explanation: There are 2 global inversions, and 1 local inversion.
solution
求逆序对数和a[i]>a[i+1]数量是否相等，可判断max(a[0:i-2])是否有大于a[i]即可
class Solution {
public:
bool isIdealPermutation(vector<int>& A) {
if (A.size() <= 2) return true;
int t = A[0];
for (int i = 2; i < A.size(); i++) {
t = max(t, A[i - 2]);
if (t > A[i]) {
return false;
}
}
return true;
}
};