为什么80%的码农都做不了架构师？>>> 
想法：用一个掩码获取除最高位之外的剩下的几位 其中掩码分别是 0，2-1，4-1，8-1，16-1，…
当mask = 7，x = 9时，ans[x&mask]是后面几位中数字1的个数
class Solution {public int[] countBits(int num) {int[] ans = new int[num+1];ans[0] = 0;int mask = 0;for(int x=1;x<=num;x++){ans[x] = 1+ans[(x&mask)];if((mask&x)==mask){mask = (mask<<1)+1;}}return ans;}}
转载于
//my.oschina.net/reter/blog/3008921
Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num ca
Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num ca
分析 打算用动态规划，所以先找x和x-1的规律： 0 1 10 11 100 101 110 111 1000 1001 10
Problem: Given a non negative integer number num. For every numbers i in the range
Question 338. Counting Bits > Question: Given a non negative integer number num. For ev
1.题目 Given a non negative integer number num. For every numbers i in the range 0 ≤ i
/\ 338. Counting Bits Given a non negative integer number num. For every numbers i i
Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num ca
[为什么80%的码农都做不了架构师？>>> ][80_] ![hot3.png][] ![c5a2642370c2ca3440abed1546dc0893e51.jpg][
试题 Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ n
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