Moderate 不用比较符返回较大的数 @CareerCup

通过判断a-b的最高位来知道哪一个数比较大

但是要注意overflow的情况

if a and b have different signs://  if a > 0, then b <  <d,  and  k = l.// if a < 0, then b > 0, and k = 0.// so either way, k = sign(a)let k = sign(a)elselet  k  = sign(a  -  b)  //  overflow  is  impossible

但因为不能用if else所以用乘法来代替: 

int k = use_sign_of_a * sa + use_sign_of_c * sc;

package Moderate;/** * Write a method which finds the maximum of two numbers. You should not use if-else or any other comparison operator.EXAMPLEInput: 5, 10Output: 10译文：写一个函数返回两个数中的较大者，你不能使用if-else及任何比较操作符。  * */public class S17_4 {/* Flips a 1 to a 0 and a 0 to a 1 */    public static int flip(int bit) {            return 1 ^ bit;    }        /* Returns 1 if a is positive, and 0 if a is negative */    public static int sign(int a) {            return flip((a >> 31) & 0x1);    }        public static int getMaxNaive(int a, int b) {            int k = sign(a - b);            int q = flip(k);            return a * k + b * q;    }        public static int getMax(int a, int b) {            int c = a - b;                        int sa = sign(a); // if a >= 0, then 1 else 0            int sb = sign(b); // if b >= 0, then 1 else 0            int sc = sign(c);// depends on whether or not a - b overflows                        /* We want to define a value k which is 1 if a > b and 0 if a < b.              * (if a = b, it doesn't matter what value k is) */                        int use_sign_of_a = sa ^ sb; // If a and b have different signs, then k = sign(a)            int use_sign_of_c = flip(sa ^ sb); // If a and b have the same sign, then k = sign(a - b)                        /* We can't use a comparison operator, but we can multiply values by 1 or 0 */            int k = use_sign_of_a * sa + use_sign_of_c * sc;            int q = flip(k); // opposite of k                        return a * k + b * q;    }        public static void main(String[] args) {            int a = 26;            int b = -15;                        System.out.println("max_naive(" + a + ", " + b + ") = " + getMaxNaive(a, b));            System.out.println("max(" + a + ", " + b + ") = " + getMax(a, b));                                        a = -15;            b = 2147483647;                        System.out.println("max_naive(" + a + ", " + b + ") = " + getMaxNaive(a, b));            System.out.println("max(" + a + ", " + b + ") = " + getMax(a, b));    }}