Task description
A binary gap within a positive integer N is any maximal sequence of consecutive zeros that is surrounded by ones at both ends in the binary representation of N.
For example, number 9 has binary representation 1001 and contains a binary gap of length 2. The number 529 has binary representation 1000010001 and contains two binary gaps: one of length 4 and one of length 3. The number 20 has binary representation 10100 and contains one binary gap of length 1. The number 15 has binary representation 1111 and has no binary gaps. The number 32 has binary representation 100000 and has no binary gaps.
Write a function:
class Solution { public int solution(int N); }
that, given a positive integer N, returns the length of its longest binary gap. The function should return 0 if N doesn’t contain a binary gap.
For example, given N = 1041 the function should return 5, because N has binary representation 10000010001 and so its longest binary gap is of length 5. Given N = 32 the function should return 0, because N has binary representation ‘100000’ and thus no binary gaps.
Write an efficient algorithm for the following assumptions:
- N is an integer within the range [1..2,147,483,647].
class Solution {
public int solution(int N) {
// write your code in Java SE 8
String a = Integer.toBinaryString(N);
char[] b = a.toCharArray();
int length = b.length;
int result = 0;
int high, low = 0;
for (int i = 0; i < length; i++) {
if (i != length - 1) {
if (b[i] != b[i + 1]) {
if ('1' == b[i]) {
low = i;
} else {
high = i + 1;
int count = high - low - 1;
if (result < count) {
result = count;
}
}
}
}
}
return result;
}
}
我的做法是先找出10的開頭並記錄1的位置，直到找出01的時候再紀錄1的位置，將兩者位置相減後再減1，得出中間有幾個0，且紀錄於一個變數中，過程中再比對出哪一次的0最多。
後來發現更簡單的方式，直接split 1 然後看陣列中哪個長度最長，後來發現這方法還是要判斷1的位置，沒有特別好。
執行結果 : https://app.codility.com/demo/results/trainingJYVTDX-7T6/
Cheng® 