（题解Java）《LeetCode零基础指南》(第二讲) 函数

直接求，做题为主；或者使用位运算；

class Solution {
    public int getSum(int a, int b) {
        return a + b;
    }
}

class Solution {
    public int getSum(int a, int b) {
         while (b != 0) {
            int carry = (a & b) << 1;
            a = a ^ b;
            b = carry;
        }
        return a;
    }
}

直接和上一题一样的做法，位运算；

class Solution {
    public int add(int a, int b) {
        while (b != 0) {
            int carry = (a & b) << 1;
            a = a ^ b;
            b = carry;
        }
        return a;
    }
}

直接上一题的做法，位运算

class Solution {
    public int add(int a, int b) {
        while (b != 0) {
            int carry = (a & b) << 1;
            a = a ^ b;
            b = carry;
        }
        return a;
    }
}

直接换成最原始的乘法原理，即加法，判断A和B两个数的大小，然后进行递归相加。

class Solution {
    public int multiply(int A, int B) {
        if(A > B)
        {
            int mid = A;
            A = B;
            B = mid;
        }
        return digui(A,B);
    }
    public int digui(int count, int B)
    {
        if(count == 0)return 0;
        return digui(count - 1, B) + B;
    }
}

有什么错误条件就用if来进行排除（主要是int负值的边界）

class Solution {
    public int divide(int dividend, int divisor) {
        if(dividend < - 2147483647 && divisor == -1 )dividend = - 2147483647;
        return dividend / divisor;
    }
}

直接计算就好了

class Solution {
    public double myPow(double x, int n) {
        return Math.pow(x,n);
    }
}

直接计算

class Solution {
    public int mySqrt(int x) {
        return (int)Math.sqrt((double)x);
    }
}

class Solution {
    public int maximum(int a, int b) {
        return Math.max(a,b);
    }
}