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[LeetCode] 868. Binary Gap #868

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grandyang opened this issue May 30, 2019 · 0 comments
Open

[LeetCode] 868. Binary Gap #868

grandyang opened this issue May 30, 2019 · 0 comments

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@grandyang
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grandyang commented May 30, 2019

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

这道题给了我们一个正整数,问其二进制表示数中两个 '1' 之间的最大距离是多少。比如整数 22 的二进制为 10110,那么可以看出前两个 '1' 之间的距离最大,所以返回2即可。其实这道题的考察点就是如何按位取数,对一个二进制数直接与 '1',就可以提取出最低位,然后和0比较大小,就知道最低位是0还是1了。当然为了每一位的是0还是1,一般有两种处理方法,一种是直接将原数字右移一位,则之前的最低位被移除了,再和 '1' 相与即可,直到原数字变为0停止。另一种方法是直接右移固定的位数再与 '1',因为整型数只有 32 位,所以可以直接取出任意位上的数字。那么既然要求两个 '1' 之间的最大距离,那么只要关心 '1' 的位置即可,一种比较直接的思路就是先遍历一遍各位上的数字,将所有 '1' 的坐标位置都保存到一个数组 pos 之中,然后再遍历这个 pos 数组,计算相邻两个数字的差,即两个 '1' 之间的距离,更新结果 res 即可得到最大距离,参见代码如下:
解法一:

class Solution {
public:
    int binaryGap(int N) {
        vector<int> pos;
        for (int i = 0; i < 32; ++i) {
            if (((N >> i) & 1) != 0) pos.push_back(i);
        }
        int res = 0, n = pos.size();
        for (int i = 1; i < n; ++i) {
            res = max(res, pos[i] - pos[i - 1]);
        }
        return res;
    }
};

我们也可以只用一个循环来完成,而且不用 pos 数组,只用一个变量 last 来记录上一个遍历到的 '1' 的位置,初始化为 -1。那么在遍历的过程中,若遇到了 '1',则判断 last 是否大于等于0,是的话则表示之前已经遇到了 '1',那么当前位置i减去 last,即为两个 '1' 之间的距离,用来更新结果 res,然后再更新 last 为当前位置i,继续遍历即可,参见代码如下:
解法二:

class Solution {
public:
    int binaryGap(int N) {
        int res = 0, last = -1;
        for (int i = 0; i < 32; ++i) {
            if (((N >> i) & 1) != 0) {
                if (last >= 0) res = max(res, i - last);
                last = i;
            }
        }
        return res;
    }
};

Github 同步地址:

#868

参考资料:

https://leetcode.com/problems/binary-gap/

https://leetcode.com/problems/binary-gap/discuss/149945/Simple-Java-(10-ms)

https://leetcode.com/problems/binary-gap/discuss/149835/C%2B%2BJavaPython-Dividing-by-2

LeetCode All in One 题目讲解汇总(持续更新中...)

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