LeetCode - Algorithms - 371. Sum of Two Integers

这题不Easy呀，像智力题，像脑筋急转弯。
代码是人家的，我可想不出来。真是“会者不难，难者不会”。

• How do I add two numbers without using any operator C/C++?
• What is the best way to add two numbers without using the + operator?
• Add two numbers without using arithmetic operators

JavaScript

自增自减运算符

/**
 * @param {number} a
 * @param {number} b
 * @return {number}
 */
var getSum = function(a, b) {
    if (a>0) {
        while(a>0) {
            a--;
            b++
        }        
    }
    if (a<0) {
        while(a<0) {
            a++;
            b--;
        }
    }
    return b;
};

Submission Detail

• 13 / 13 test cases passed.
• Your runtime beats 3.38 % of javascript submissions.

位运算+递归

/**
 * @param {number} a
 * @param {number} b
 * @return {number}
 */
var getSum = function(a, b) {
	  var sum,carry;
	  if (b==0) return a;
	  sum = a^b;
	  carry = (a&b)<<1;
	  return getSum(sum,carry);
};

Submission Detail

• 13 / 13 test cases passed.
• Your runtime beats 69.57 % of javascript submissions

消除递归的位运算

知其然而不知其所以然，居然可以这么做，觉得很神奇

/**
 * @param {number} a
 * @param {number} b
 * @return {number}
 */
var getSum = function(a, b) {
	  while(b!=0) {
		  var carry=a&b;
		  a = a^b;
		  b = carry<<1;
	  }
	  return a;
};

Submission Detail

• 13 / 13 test cases passed.
• Your runtime beats 100.00 % of javascript submissions.

Java

消除递归的位运算

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

Submission Detail

• 13 / 13 test cases passed.
• Your runtime beats 100.00 % of java submissions.

递归+位运算

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

Submission Detail

• 13 / 13 test cases passed.
• Your runtime beats 100.00 % of java submissions.

自增自减运算符

这个解很像脑筋急转弯，不难，但让人想不到

class Solution {
    public int getSum(int a, int b) {
    	if (a>0) {
    		while(a>0) {
    			a--;
    			b++;
    		}
    	}
    	if (a<0) {
    		while(a<0) {
    			a++;
    			b--;
    		}
    	}
    	return b;  
    }
}

Submission Detail

• 13 / 13 test cases passed.
• Your runtime beats 3.44 % of java submissions.