You are given an integer array?nums?of length?n?which represents a permutation of all the integers in the range?[0, n - 1].

The number of?global inversions?is the number of the different pairs?(i, j)?where:

• 0 <= i < j < n

• nums[i] > nums[j]

The number of?local inversions?is the number of indices?i?where:

• 0 <= i < n - 1

• nums[i] > nums[i + 1]

Return?true?if the number of?global inversions?is equal to the number of?local inversions.

?

Example 1:

Input: nums = [1,0,2]

Output: true

Explanation: There is 1 global inversion and 1 local inversion.

Example 2:

Input: nums = [1,2,0]

Output: false

Explanation: There are 2 global inversions and 1 local inversion.

這里面local的就一定是global的，所以如果要返回false就是當(dāng)存在num[i]>num[j];

同時i+2<=j;

我們保存一個max，讓max去跟目前的j去比對，即可；

?

Constraints:

• n == nums.length

• 1 <= n <= 105

• 0 <= nums[i] < n

• All the integers of?nums?are?unique.

• nums?is a permutation of all the numbers in the range?[0, n - 1].

Runtime:?1 ms, faster than?100.00%?of?Java?online submissions for?Global and Local Inversions.

Memory Usage:?51.6 MB, less than?70.61%?of?Java?online submissions for?Global and Local Inversions.