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 1:
Input: 2 Output: [0,1,1]
Example 2:
Input: 5
Output: [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.
给定一个非负整数 num。对于 0 ≤ i ≤ num 范围中的每个数字 i ，计算其二进制数中的 1 的数目并将它们作为数组返回。
示例 1:
输入: 2 输出: [0,1,1]
示例 2:
输入: 5
输出: [0,1,1,2,1,2]
进阶:
- 给出时间复杂度为O(n*sizeof(integer))的解答非常容易。但你可以在线性时间O(n)内用一趟扫描做到吗？
- 要求算法的空间复杂度为O(n)。
- 你能进一步完善解法吗？要求在C++或任何其他语言中不使用任何内置函数（如 C++ 中的 __builtin_popcount）来执行此操作。
36ms
1 class Solution {
2 func countBits(_ num: Int) -> [Int] {
3 guard num > 0 else {
4 return [0]
5 }
6
7 var result = [0]
8 var i = 0
9 var total = 1
10
11 for j in 1...num {
12 result.append(result[i] + 1)
13 i += 1
14 if i == total {
15 i = 0
16 total = j + 1
17 }
18 }
19 return result
20 }
21 }
40ms
1 class Solution {
2 func countBits(_ num: Int) -> [Int] {
3 if num == 0 {
4 return [0]
5 }
6 var results = [Int]()
7 results.append(0)
8 for n in 1...num {
9 if n%2 == 1 {
10 results.append(results[n-1]+1)
11 } else {
12 results.append(results[n/2])
13 }
14 }
15 return results
16 }
17 }
44ms
1 class Solution {
2 func countBits(_ num: Int) -> [Int] {
3 guard num > 0 else {
4 return [0]
5 }
6
7 var result = Array(repeating: 0, count: num + 1)
8 result[1] = 1
9 var loopCount = 2
10 var index = 2
11 while index <= num {
12 for j in 0..<loopCount {
13 if index > num {
14 break
15 }
16
17 result[index] = 1 + result[j]
18 index += 1
19 }
20
21 loopCount *= 2
22 }
23 return result
24 }
25 }
48ms
1 class Solution {
2 func countBits(_ num: Int) -> [Int] {
3 var a = [0]
4 var b = [Int]()
5 while (a.count + b.count) != num+1 {
6 if a.count == b.count{
7 a = a + b
8 b = [Int]()
9 }
10 b.append(a[b.count]+1)
11 }
12 return a + b
13 }
14 }
88ms
1 class Solution {
2 func countBits(_ num: Int) -> [Int] {
3 guard num > 0 else { return [0] }
4 var dp = [Int](repeating: 0, count: num + 1)
5
6 for i in 1...num {
7 dp[i] = dp[i & (i - 1)] + 1
8 }
9 return dp
10 }
11 }
108ms
1 class Solution {
2 func countBits(_ num: Int) -> [Int] {
3 var result = [Int]()
4 if num < 0 {
5 return result
6 }
7 result.append(0)
8 if num == 0 {
9 return result
10 }
11 for i in 1 ... num {
12 print(i & 1)
13 result.append(result[i >> 1] + (i & 1))
14 }
15 return result
16 }
17 }
112ms
1 class Solution {
2 func countBits(_ num: Int) -> [Int] {
3 if num == 0 { return [0] }
4 var result = [0]
5 var count = 1
6 while true {
7 for j in 0 ..< count {
8 result.append(result[j] + 1)
9 if result.count == num + 1 {
10 return result
11 }
12 }
13 count = result.count
14 }
15 return result
16
17 }
18 }
116ms
1 class Solution {
2 func countBits(_ num: Int) -> [Int] {
3 if num == 0 {
4 return [0]
5 }
6 var res = [Int].init()
7 res.append(0)
8 for n in 1...num {
9 let count = res[n & (n - 1)] + 1
10 res.append(count)
11 }
12 return res
13 }
14 }
164ms
1 class Solution {
2 func countBits(_ num: Int) -> [Int] {
3 var b = [Int](repeating: 0, count: num + 1)
4 if num == 0 { return [0] }
5 if num == 1 { return [0,1] }
6 for i in 1...num {
7 b[i] = b[i >> 1] + i % 2
8 }
9 return b
10 }
11 }
176ms
1 class Solution {
2 func countBits(_ num: Int) -> [Int]
3 {
4 var res = [Int]()
5 for i in 0...num
6 {
7 var n = i
8 var count = 0
9 while n > 0
10 {
11 if n&1 == 1
12 {
13 count += 1
14 }
15 n = n >> 1
16 }
17 res.append(count)
18 }
19 return res
20 }
21 }
156ms
1 class Solution {
2 func countBits(_ num: Int) -> [Int] {
3 guard num > 0 else { return [0] }
4 var dp = [Int](repeating: 0, count: num + 1)
5
6 for i in 1...num {
7 dp[i] = dp[i / 2]
8 if i % 2 == 1 {
9 dp[i] += 1
10 }
11 }
12 return dp
13 }
14 }
原文地址：https://www.cnblogs.com/strengthen/p/10262072.html
时间： 2024-10-04 00:03:15