[leetcode] (周赛)869. 重新排序得到 2 的幂

869. 重新排序得到 2 的幂

枚举排列，然后验证。比较暴力。

其实好一点的做法应该反过来，先把int范围下的2的N幂算出来，然后一个一个验证给出的数能不能拼成。

class Solution {
    public Integer[] nextPermutation(Integer[] nums) {
        if (nums.length == 1) {
            return new Integer[]{};
        }
        int p = -1;
        for (int i = nums.length - 2; i >= 0; i--) {
            if (nums[i] < nums[i + 1]) {
                p = i;
                break;
            }
        }

        if (p != -1) {
            int tmp = nums[p];
            int q = nums.length - 1;
            while (nums[q] <= tmp) {
                q--;
            }

            nums[p] = nums[q];
            nums[q] = tmp;

            reverse(p + 1, nums);
        } else {
            return new Integer[]{};
        }

        return nums;
    }

    public void reverse(int k, Integer[] nums) {
        if (k >= nums.length) return;
        int i = k;
        int j = nums.length - 1;
        while (i < j) {
            int tmp = nums[i];
            nums[i] = nums[j];
            nums[j] = tmp;
            i++;
            j--;
        }
    }

    public boolean reorderedPowerOf2(int N) {
        if (N == 0) {
            return false;
        }

        String s = Integer.toString(N);
        int[] p = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
        for (int i = 0; i < s.length(); i++) {
            p[s.charAt(i) - '0']++;
        }

        List<Integer> list = new ArrayList<>();

        for (int i = 0; i < 10; i++) {
            for (int j = 0; j < p[i]; j++) {
                list.add(i);
            }
        }

        Integer[] nums = list.stream().toArray(Integer[]::new);

        StringBuilder origin = new StringBuilder();
        for (int i = 0; i < nums.length; i++) {
            origin.append(nums[i]);
        }

        while (nums[0] == 0) {
            nums = nextPermutation(nums);
        }

        int flag = 0;

        while (true) {
            if (nums.length == 0) {
                break;
            }

            StringBuilder st = new StringBuilder();
            for (int i = 0; i < nums.length; i++) {
                st.append(nums[i]);
            }

            Integer num = Integer.valueOf(st.toString());

            double ans = Math.log(num) / Math.log(2);
            if (isNotDouble(ans)) {
                return true;
            }
            nums = nextPermutation(nums);
        }

        return false;
    }

    public static boolean isNotDouble(Double x) {
        int i = x.intValue();
        return x - i < 0.000001;
    }

}

这周周赛就做出来这两个。。。水平还是太低啊，哎