Saturday, March 19, 2016

LeetCode 338 - Counting Bits


https://www.hrwhisper.me/leetcode-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.

想一想,当一个数为2的整数幂的时候,1的个数为1,比如2(10) 和4(100),8(1000)
在这之后就是前一个序列的数+1 比如 9(1001) = 1(1) + 8 (1) = 2
就是把一个数分解为小于它的最大2的整数幂 + x
    public int[] countBits(int num) {
        int[] res = new int[num+1];
        int pow2 = 1,before =1;
        for(int i=1;i<=num;i++){
            if (i == pow2){
                before = res[i] = 1;
                pow2 <<= 1;
            }
            else{
                res[i] = res[before] + 1;
                before += 1;
            }
        }
        return res;
    }

https://leetcode.com/discuss/92609/three-line-java-solution
An easy recurrence for this problem is f[i] = f[i / 2] + i % 2.
public int[] countBits(int num) {
    int[] f = new int[num + 1];
    for (int i=1; i<=num; i++) f[i] = f[i >> 1] + (i & 1);
    return f;
}
倒过来想,一个数 * 2 就是把它的二进制全部左移一位,也就是说 1的个数是相等的。
那么我们可以利用这个结论来做。
res[i /2] 然后看看最低位是否为1即可(上面*2一定是偶数,这边比如15和14除以2都是7,但是15时通过7左移一位并且+1得到,14则是直接左移)
所以res[i] = res[i >>1] + (i&1)
    public int[] countBits(int num) {
        int[] res = new int[num+1];
        for(int i=1;i<=num;i++){
            res[i] = res[i >> 1] + (i & 1);
        }
        return res;

https://leetcode.com/discuss/92609/three-line-java-solution
public int[] countBits(int num) { int[] answer = new int[num+1]; int offset = 1; for(int i = 1; i < answer.length; i++){ if(offset * 2 == i) offset *= 2; answer[i] = 1 + answer[i - offset]; } return answer; }
https://leetcode.com/discuss/93807/simple-java-dynamic-programming-without-bitwise-operation
int[] bits = new int[num + 1]; for(int i = 1; i <= num; i++){ bits[i] = bits[i/2]; if(i%2 == 1) bits[i]++; } return bits; }
Brute Force:
http://blog.csdn.net/lnho2015/article/details/50924299
public int[] countBits(int num) { int[] result=new int[num+1]; result[0]=0; for(int i=1;i<=num;i++){ result[i]=getCount(i); } return result; } public int getCount(int num){ int count=0; while(num!=0){ if((num&1)==1){ count++; } num/=2; } return count; }
https://leetcode.com/discuss/92675/handle-this-question-interview-thinking-process-solution

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