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_338.java
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package com.fishercoder.solutions;
import com.fishercoder.common.utils.CommonUtils;
/**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?
*
*
*/
public class _338 {
private class DPSolution {
//lixx2100's post is cool:https://discuss.leetcode.com/topic/40162/three-line-java-solution
//An easy recurrence for this problem is f[i] = f[i / 2] + i % 2
//and then we'll use bit manipulation to express the above recursion function
// right shift by 1 means to divide by 2
//AND with 1 means to modulo 2
//this is so cool!
public int[] countBits(int num) {
int[] ones = new int[num + 1];
for (int i = 1; i <= num; i++) {
ones[i] = ones[i >> 1] + (i & 1);
}
return ones;
}
}
//use the most regular method to get it AC'ed first
public int[] countBits(int num) {
int[] ones = new int[num + 1];
for (int i = 0; i <= num; i++) {
ones[i] = countOnes(i);
}
return ones;
}
private int countOnes(int i) {
int ones = 0;
while (i != 0) {
ones++;
i &= (i - 1);
}
return ones;
}
public static void main(String... strings) {
_338 test = new _338();
int num = 15;
int[] ones = test.countBits(num);
CommonUtils.printArray(ones);
}
}