Description
Difficulty：Easy
Tag： Math
Given a positive integer N, find and return the longest distance between two consecutive 1's in the binary representation of N.
If there aren't two consecutive 1's, return 0.
Example 1:
Input: 22
Output: 2
Explanation:
22 in binary is 0b10110.
In the binary representation of 22, there are three ones, and two consecutive pairs of 1's.
The first consecutive pair of 1's have distance 2.
The second consecutive pair of 1's have distance 1.
The answer is the largest of these two distances, which is 2.
Example 2:
Input: 5
Output: 2
Explanation:
5 in binary is 0b101.
Example 3:
Input: 6
Output: 1
Explanation:
6 in binary is 0b110.
Example 4:
Input: 8
Output: 0
Explanation:
8 in binary is 0b1000.
There aren't any consecutive pairs of 1's in the binary representation of 8, so we return 0.
Note:
1 <= N <= 10^9
Solution
其實是一道數1的題,用 two pointers 解決
func binaryGap(N int) int {
max,pre,next := 0,0,1
for ;N & 1 == 0 && N > 0; pre,N = pre+1, N >> 1{
}
for next,N = pre + 1,N>>1;N > 0; N>>=1 {
for N & 1 == 0{
next++
N >>=1
}
if next -pre > max {
max = next - pre
}
pre = next
next += 1
}
return max
}
網友有一個更精妙簡單的解法：
public int binaryGap(int N) {
int res = 0;
for (int d = -32; N > 0; N /= 2, d++)
if (N % 2 == 1) {
res = Math.max(res, d);
d = 0;
}
return res;
}
d 是兩個1之間距離，碰到1之後d置0.