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LeetCode 338. Counting Bits
Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array.
Example:
For num = 5
you should return [0,1,1,2,1,2]
.
Follow up:
It is very easy to come up with a solution with run time O(n*sizeof(integer)). But can you do it in linear time O(n) /possibly in a single pass?
Space complexity should be O(n).
Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.
Hint:
You should make use of what you have produced already.
Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.
Or does the odd/even status of the number help you in calculating the number of 1s?
给定一个非负整数num。对于每一个满足0 ≤ i ≤ num的数字i,计算其数字的二进制表示中1的个数,并以数组形式返回。
测试用例如题目描述。
进一步思考:
很容易想到运行时间 O(n*sizeof(integer)) 的解法。但你可以用线性时间O(n)的一趟算法完成吗?
空间复杂度应当为O(n)。
你可以像老板那样吗?不要使用任何内建函数(比如C++的__builtin_popcount)。
提示:
你应当利用已经生成的结果。
将数字拆分为诸如 [2-3], [4-7], [8-15] 之类的范围。并且尝试根据已经生成的范围产生新的范围。
3. 数字的奇偶性可以帮助你计算1的个数吗?
解法I 利用移位运算:
递推式:ans[n] = ans[n >> 1] + (n & 1)
//c++版本 class Solution { public: vector<int>countBits(int num) { //一个数组有(0~num)即num+1个元素,初始化为0 vector<int> v1(num+1,0); for(int i=1;i<=num;i++) { v1[i]=v1[i>>1]+(i&1); } } }
15 / 15 test cases passed.
Status: Accepted
Runtime: 124 ms
Submitted: 0 minutes ago
解法II 利用highbits运算:
递推式:ans[n] = ans[n - highbits(n)] + 1
其中highbits(n)
表示只保留n的最高位得到的数字。
highbits(n) = 1<<int(math.log(x,2)) math.log()不是c++/c的函数,java中有
例如:
highbits(7) = 4 (7的二进制形式为111) highbits(10) = 8 (10的二进制形式为1010)
解法III 利用按位与运算:
递推式:ans[n] = ans[n & (n - 1)] + 1
//c++版本 class Solution { public: vector<int>countBits(int num) { //一个数组有(0~num)即num+1个元素,初始化为0 vector<int> v1(num+1,0); for(int i=1;i<=num;i++) { v1[i]=v1[n&(n-1)]+1; } } }
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