return ((a - b) >> 31) & 1 == 1?b:a;

if (((a - b) >> 31) & 1 == 1){
    return b;
}
return a;

public int flip(int n) {
    return n ^ 1;
}

    public static int sign(int n) {
        return flip((n >> 31) & 1);
    }

sign(a - b) * a + flip(sign(a - b)) * b

public static int getMax1(int a, int b) {
    int c = a - b;
    //當(dāng)符號(hào)位是 1 的時(shí)候，scA = 0，符號(hào)位是 0 的時(shí)候，scA = 1。
    int scA = sign(c);
    // scA = 1 時(shí)，scB = 0，scA = 0時(shí)，scB = 1
    int scB = flip(scA);
    // 如果 scA = 0，說(shuō)明 b 大，直接返回b
    // 如果 scA = 1，說(shuō)明 a 大，直接返回a
    return a * scA + b * scB;
}

a = 2147483647;
b = -2147480000;

int c = a - b;
int sa = sign(a);
int sb = sign(b);
int sc = sign(c);
int difSab = sa ^ sb;
int sameSab = flip(difSab);
int returnA = difSab * sa + sameSab * sc;
int returnB = flip(returnA);
return a * returnA + b * returnB;

// 不要用任何比較判斷，返回兩個(gè)數(shù)中較大的數(shù)
public class Code_GetMax {

    public static int flip(int n) {
        return n ^ 1;
    }


    public static int sign(int n) {
        return flip((n >> 31) & 1);
    }

    public static int getMax1(int a, int b) {
        int c = a - b;
        int scA = sign(c);
        int scB = flip(scA);
        return a * scA + b * scB;
    }

    public static int getMax2(int a, int b) {
        int c = a - b;
        int sa = sign(a);
        int sb = sign(b);
        int sc = sign(c);
        int difSab = sa ^ sb;
        int sameSab = flip(difSab);
        int returnA = difSab * sa + sameSab * sc;
        int returnB = flip(returnA);
        return a * returnA + b * returnB;
    }

    public static void main(String[] args) {
        int a = -16;
        int b = -19;
        System.out.println(getMax1(a, b));
        System.out.println(getMax2(a, b));
        a = 2147483647;
        b = -2147480000;
        System.out.println(getMax1(a, b)); // wrong answer because of overflow
        System.out.println(getMax2(a, b));

    }
}