869. 重新排序得到 2 的幂

给定正整数 N ，我们按任何顺序（包括原始顺序）将数字重新排序，注意其前导数字不能为零。

如果我们可以通过上述方式得到 2 的幂，返回 true；否则，返回 false。

示例 1：

输入：1
输出：true
示例 2：

输入：10
输出：false
示例 3：

输入：16
输出：true
示例 4：

输入：24
输出：false
示例 5：

输入：46
输出：true

---

class Solution {

    public boolean reorderedPowerOf2(int N) {
        // Build eg. N = 128 -> A = [1, 2, 8]
        String S = Integer.toString(N);
        int[] A = new int[S.length()];
        for (int i = 0; i < S.length(); ++i) {
            A[i] = S.charAt(i) - '0';
        }
        return permutations(A, 0);
    }

    // Return true if A represents a valid power of 2
    public boolean isPowerOfTwo(int[] A) {
        if (A[0] == 0) {
            return false;  // no leading zero
        }

        // Build eg. A = [1, 2, 8] -> N = 128
        int N = 0;
        for (int x : A) {
            N = 10 * N + x;
        }

        

        // Check that there are no other factors besides 2
        return (N & (N-1))==0;
    }

    /**
     * Returns true if some permutation of (A[start], A[start+1], ...) can result in A representing
     * a power of 2.
     */
    public boolean permutations(int[] A, int start) {
        if (start == A.length) {
            return isPowerOfTwo(A);
        }

        // Choose some index i from [start, A.length - 1]
        // to be placed into position A[start].
        for (int i = start; i < A.length; ++i) {
            // Place A[start] with value A[i].
            swap(A, start, i);

            // For each such placement of A[start], if a permutation
            // of (A[start+1], A[start+2], ...) can result in A
            // representing a power of 2, return true.
            if (permutations(A, start + 1)) {
                return true;
            }

            // Restore the array to the state it was in before
            // A[start] was placed with value A[i].
            swap(A, start, i);
        }

        return false;
    }

    public void swap(int[] A, int i, int j) {
        int t = A[i];
        A[i] = A[j];
        A[j] = t;
    }
}