public int lastRemaining(int n) {
return leftToRight(n);
}
private static int leftToRight(int n) {
if(n <= 2) return n;
return 2 * rightToLeft(n / 2);
}
private static int rightToLeft(int n) {
if(n <= 2) return 1;
if(n % 2 == 1) return 2 * leftToRight(n / 2);
return 2 * leftToRight(n / 2)  1;
}
O(logN) solution. clear break down



Clever use of mutual recursion, almost selfexplained! I reduce the base case a little bit and add some comments for easier understand.
public int lastRemaining(int n) { return leftToRight(n); } // eliminate [1...n] first from left to right, then alternate private int leftToRight(int n) { if (n == 1) return 1; // scan from left to right is simple, the length of array doesn't matter // [1, 2, 3, 4] > 2 * [1, 2] // [1, 2, 3, 4, 5] > 2 * [1, 2] return 2 * rightToLeft(n / 2); } // eliminate [1...n] first from right to left, then alternate private int rightToLeft(int n) { if (n == 1) return 1; // if the length of array is even, we will get only odd number // [1, 2, 3, 4] > [1, 3] = 2 * [1, 2]  1 if (n % 2 == 0) return 2 * leftToRight(n / 2)  1; // else if the length of array is odd, we will get only even number // [1, 2, 3, 4, 5] > [2, 4] = 2 * [1, 2] else return 2 * leftToRight(n / 2); }