Recover Rotated Sorted Array

Problem: Given a rotated sorted array, recover the sorted array in-place.

For example:
{ 4, 5, 1, 2, 3 } –> {1, 2, 3, 4, 5}

Challenge requirement:
in-palce, O(1) extra space and O(n) time

Solution:
3-step reverse operation
Step 1: find two different sub array sections {4, 5}, {1, 2, 3}
Step 2. reverse sub array elements to: {5, 4}, { 3, 2, 1}
Step 3: reverse all array elements to {1, 2, 3, 4, 5}

Code:


	public void recoverRotatedSortedArray(ArrayList<Integer> nums) {
		if(nums.size() <= 1) { return; } // do nothing
		if(nums.get(0) < nums.get(nums.size() - 1)) {  // do nothing
			return;
		}
		
		int pt = 0;
		for(int i = 0; i < nums.size() - 1; i++) {
			if(nums.get(i) > nums.get(i+1)) {
				pt = i;
			}
		}

		for(int start = 0, end = pt; start < end; start++, end--) {
			swap(nums, start, end);
		}
		
		for(int start = pt+1, end = nums.size() - 1; start < end; start++, end--) {
			swap(nums, start, end);
		}
		
		for(int start = 0, end = nums.size() - 1; start < end; start++, end--) {
			swap(nums, start, end);
		}
		
	}


	private void swap(ArrayList<Integer> nums, int start, int end) {
		int tmp = nums.get(start);
		nums.set(start, nums.get(end));
		nums.set(end, tmp);
	}