[LeetCode]600. Non-negative Integers without Consecutive Ones

https://leetcode.com/problems/non-negative-integers-without-consecutive-ones/#/description

给一个数n，找出有多少个不大于n的正数，且这些正数的二进制表示中不包含两个连续的1

int转2进制string的API！！

先用dp找出在k位二进制长度的数有多少满足条件的。a[i]表示以0为结尾的长度为i + 1的满足条件的数字个数，b[i]表示1为结尾的。

再找a[n - 1] + b[n - 1]里面overcount了多少，只有当连续两位为0时，b[i]是overcount的。11、01、10都能cover到a[i]和b[i]

public class Solution {    public int findIntegers(int num) {        StringBuilder sb = new StringBuilder(Integer.toBinaryString(num)).reverse();        int n = sb.length();        int[] a = new int[n];        int[] b = new int[n];        a[0] = 1;        b[0] = 1;        for (int i = 1; i < n; i++) {            a[i] = a[i - 1] + b[i - 1];            b[i] = a[i - 1];        }        int res = a[n - 1] + b[n - 1];        for (int i = n - 2; i >= 0; i--) {            if (sb.charAt(i) == '0' && sb.charAt(i + 1) == '0') {                res -= b[i];            }            if (sb.charAt(i) == '1' && sb.charAt(i + 1) == '1') {                break;            }        }        return res;    }}