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