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191. Number of 1 Bits
Write a function that takes an unsigned integer and returns the number of ’1‘ bits it has (also known as theHamming weight).
For example, the 32-bit integer ’11‘ has binary representation 00000000000000000000000000001011, so the function should return 3.
Solution 1:
看起来是一道非常简单的题目，但是自己在做的时候却漏洞百出。
思路：存一个count，每次将n向右移一位，检查最后一位是否为1，是1就将count加一。
注意shift number不会改变number本身，要赋值给number本身才可以。
注意overflow的情况，所以不能把n放在while的条件里面做判断。
Bit Manipulation Ref:
http://www.tutorialspoint.com/java/java_basic_operators.htm
public class Solution { // you need to treat n as an unsigned value public int hammingWeight(int n) { int count=31; int res=0; while(count>=0) { if((n&1)!=0) { res++; } n=n>>>1; count--; } return res; }}
Solution2:
看了discussion，居然有种作弊的方法！！！
return Integer.bitCount(n);Solution3:
public class Solution { // you need to treat n as an unsigned value public int hammingWeight(int n) { int count = 0; int mask = 1; int i = 0; while(i < 32) { if ((n&mask) != 0) { count++; } mask <<= 1; i++; } return count; }}
思路跟1是一样的，但是这种是每次移i次，。移位的简写方式：<<=,>>=. >>>三个箭头是fill left by 0.
Solution 4:
public class Solution { // you need to treat n as an unsigned value public int hammingWeight(int n) { int count = 0; for (count = 0; n != 0; count++) { n &= n - 1; // clear the least significant bit set } return count; }}
这解法太巧了！把最右边的1清零，有多少1就清零多少次。然后累加次数
111000 & 110111＝110000
110000 & 101111 ＝100000
100000 & 011111 ＝000000
191. Number of 1 Bits
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