[Offer] [15] [1] binary number

• Title Description
• Ideas analysis
• Test Case
• Code link

Title Description

　　An integer that indicates the number of the output of the binary number 1 in. Wherein a negative number indicates a complement.

Ideas analysis

1. Let integer and 1phase can be judged integer binary representation of whether the rightmost one 1, the one left after another and integer phase , you can compare the second from the right, comparing the cycle, you can count the binary 1number
   (may generally think of an integer to the right , and then after 1phase, but this approach will fall into an infinite loop when incoming negative, because negative right shift operation will be supplemented at a high level 1, instead 0)
2. After a integer minus 1 and then the original integer phase , then the binary integer in the rightmost 1 becomes 0, until the whole numbers 0, you can count the number of 1s.

Test Case

1. Positive (inclusive 1,0x7FFFFFFF)
2. Negative (inclusive 0x80000000,0xFFFFFFFF)
3. 0
   　　
   ## the Java Code

public class Offer15 {
    public static void main(String[] args) {
        System.out.println("测试正数-->");
        test1();
        System.out.println("测试负数-->");
        test2();
        System.out.println("测试0-->");
        test3();
    }
    public static int NumberOf1(int n) {
        return Solution3(n);
    }
    /**
     * 解法一： 思路： 上面的思路是移动 整数，我们可以移动与整数相与的1 ，每次判断之后我们将1 向左移动， 这时判断的就是整数所表示二进制中右边的第二位
     * 
     * @param n
     * @return
     */
    private static int Solution2(int n) {
        int count = 0;
        int flag = 1;
        while (flag != 0) {
            if ((flag & n) != 0) {
                count++;
            }
            flag = flag << 1;
        }
        return count;
    }

    /**
     * 解法二： 思路 ：利用了一种二进制运算的规律，把一个整数减去1之后再和原来的整数做与运算， 得到的结果相当于把整数二进制中最右边的1变为0
     * 
     * @param n
     * @return
     */
    private static int Solution3(int n) {
        int count = 0;
        while (n != 0) {
            count++;
            n = n & (n - 1);
        }
        return count;
    }

    /**
     * 测试正数
     */
    private static void test1() {
        System.out.println("0x7FFFFFFF: " + NumberOf1(0x7FFFFFFF));
        System.out.println("1: " + NumberOf1(1));
    }

    /**
     * 测试负数
     */
    private static void test2() {
        System.out.println("0x80000000： " + NumberOf1(0x80000000));
        System.out.println("0xFFFFFFFF： " + NumberOf1(0xFFFFFFFF));
    }

    /**
     * 测试0
     */
    private static void test3() {
        System.out.println("0: " + NumberOf1(0));
    }
}

Code link

Offer to prove safety codes -Java