[LeetCode] 338. Counting Bits
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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.
思路:
1.空间换时间,先计算出2^0 ~ 2^32 放到数组a待用。
2.将数n与数组a从大到小对比,大于某个a[i]则标记一次,并且n=n-a[i],直到对比完毕,返回标记的总次数。
3.按要求循环执行第二步,并将标记次数用数组形式输出。
Java代码:
public class Solution { public int[] countBits(int num) { int[] c = new int[num+1]; long[] preNums=getPreNums(); for(int i = 0; i <= num ; i++){ c[i] = getOne(i,preNums); System.out.println(c[i]); } return c; } public static int getOne(int num,long[] preNums){ long temp = num; int count = 0; for(int i = 32; temp > 0; i-- ){ if(temp >= preNums[i]){ temp = temp - preNums[i]; count++; } } return count; }public static long[] getPreNums(){ long[] a = new long[33]; a[0] = 1; for(int i = 1; i <= 32; i++){ a[i] = a[i-1] * 2; } return a;}}
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