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.
Note:
-
A
will be a permutation of[0, 1, ..., A.length - 1]
. -
A
will have length in range[1, 5000]
. - The time limit for this problem has been reduced.
思路:
Key insights:
- every local inversion is also a global inversion
- so “local inversions == global inversions” can be interpreted as “there are only local inversions”
- if there are only local inversions, the array will be sorted after making all local inversions
- if there are inversions that are not local, the array won’t be sorted after making all local inversions
class Solution {
public boolean isIdealPermutation(int[] A) {
for(int i = 0;i < A.length-1;){
if(A[i]>A[i+1]){
int temp = A[i];
A[i] =A[i+1];
A[i+1] = temp;
i += 2;
}else{
i++;
}
}
for(int i = 0;i < A.length -1; i++){
if(A[i]>A[i+1]){
return false;
}
}
return true;
}
} public class Main {
public static void main(String[] args){
Solution solution = new Solution();
int[] A = {1,2,0};
solution.isIdealPermutation(A);
}
}