338 Counting Bits

Author: int32bit

README.md

@@ -138,6 +138,7 @@
 + [237 Delete Node in a Linked List(O(1)删除单链表节点)](algorithms/DeleteNodeinaLinkedList)
 + [238 Product of Array Except Self(数组操作)](algorithms/ProductofArrayExceptSelf)
 + [242 Valid Anagram(hash表，判断两个字符串出现的字符和次数均相同)](algorithms/ValidAnagram)
++ [338 Counting Bits(二进制1的个数，迭代法)](algorithms/CountingBits)
 
 ## Database
 

algorithms/CountingBits/README.md

@@ -0,0 +1,52 @@
+## 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.
+
+## Solution
+
+很容易实现O(n * sizeof(num))的方法：
+
+```c
+int countBit(int n)
+{
+	int sum = 0;
+	while (n) {
+		sum++;
+		n &= (n - 1);
+	}
+	return sum;
+}
+int *countBits(int num, int *size)
+{
+	*size = num + 1;
+	int* ans = (int *)malloc(sizeof(int) * *size);
+	for (int i = 0; i <= num; ++i)
+		ans[i] = countBit(i);
+	return ans;
+}
+```
+
+但题目要求O(n),我们可以使用迭代的方法，我们知道左移位除了最后一个1可能丢掉，不会改变其余位的1的个数，假设f(n)表示n的二进制1的个数，则：
+
+* 若n为偶数，则f(n) = f(n >> 1).
+* 若n为奇数，则f(n) = f (n >> 1) + 1. 其中后面的1是由于移位丢掉的。
+
+```c
+int *countBits(int num, int *size)
+{
+	*size = num + 1;
+	int* ans = (int *)malloc(sizeof(int) * *size);
+	for (int i = 1; i <= num; ++i)
+		ans[i] = ans[i >> 1] + (i & 1);
+	return ans;
+}
+```

algorithms/CountingBits/solve.cpp

@@ -0,0 +1,50 @@
+#include <stdio.h>
+#include <stdlib.h>
+class Solution {
+	public:
+		vector<int> countBits(int num) {
+			vector<int> result(num + 1, 0);
+			for (int i = 1; i <= num; ++i)
+				result[i] = result[i >> 1] + (i & 1);
+			return result;
+		}
+};
+int countBit(int n)
+{
+	int sum = 0;
+	while (n) {
+		sum++;
+		n &= (n - 1);
+	}
+	return sum;
+}
+int *countBits_bad(int num, int *size)
+{
+	*size = num + 1;
+	int* ans = (int *)malloc(sizeof(int) * *size);
+	for (int i = 0; i <= num; ++i)
+		ans[i] = countBit(i);
+	return ans;
+}
+int *countBits(int num, int *size)
+{
+	*size = num + 1;
+	int* ans = (int *)malloc(sizeof(int) * *size);
+	for (int i = 1; i <= num; ++i)
+		ans[i] = ans[i >> 1] + (i & 1);
+	return ans;
+}
+int main(int argc, char **argv)
+{
+	int n;
+	while (scanf("%d", &n) != EOF) {
+		int size = 0;
+		int *ans = countBits(n, &size); 
+		for (int i = 0; i < size; ++i) {
+			printf("%d ", ans[i]);
+		}
+		printf("\n");
+		free(ans);
+	}
+	return 0;
+}