Divide Two Integers

2024年5月30日 138次阅读

Given two integers dividend and divisor, divide two integers without using multiplication, division and mod operator.

Return the quotient after dividing dividend by divisor.

The integer division should truncate toward zero.

Example 1:

Input: dividend = 10, divisor = 3
Output: 3

Example 2:

Input: dividend = 7, divisor = -3
Output: -2

Note:

  • Both dividend and divisor will be 32-bit signed integers.
  • The divisor will never be 0.
  • Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231,  231 − 1]. For the purpose of this problem, assume that your function returns 231 − 1 when the division result overflows.

大概意思，不要用* / 取模 这三个符号来计算除法

boolean isMinedNeg = false;
		boolean isMinsorNeg = false;
		boolean isNeg = false;
		if (dividend < 0) {
			isNeg = true;
			if (dividend == Integer.MIN_VALUE) {
				dividend = Integer.MAX_VALUE;
				isMinedNeg = true;
			} else {
				dividend = 0 - dividend;
			}
		}

		if (divisor < 0) {
			isNeg = !isNeg;
			if (divisor == Integer.MIN_VALUE) {
				divisor = Integer.MAX_VALUE;
				isMinsorNeg = true;
			} else {
				divisor = 0 - divisor;
			}
		}

		if(dividend < divisor || (isMinsorNeg && !isMinedNeg)) {
			return 0;
		}
		
		int oldtemp = 0;
		int oldcount = 0;
		int sum = 0;
		int mod = dividend - oldtemp;
		
		while(mod >= divisor) {
			int count = 1;
			int temp = divisor;
			while (mod >= temp) {
				if(temp < 0) {
					break;
				}
				oldtemp = temp;
				oldcount = count;
				temp += temp;
				count = count<<1;
			}
			mod = mod - oldtemp;
			sum += oldcount;
		}
		
		if(isNeg) {
			sum = 0 - sum;
			if(isMinedNeg && 1 + mod >= divisor) {
				sum -= 1;
			}
		}
		
		return sum;

我的大概思路，负数的全变成正数，不断的用1的两倍来解决效率的问题，举一个例子：60/3  1.3+3=6 <= 60     1*2=2    6+6=12 <= 60    2*2 =4    12+12=24 <= 60  4*2=8     24+24=48 <= 60  8*2=16     48+48=96 > 60 中断 2.因为60 -48 =12    3+3=6 <=
12     1*2=2    6+6=12 <= 12   2*2=4    12+12=24 > 12 中断 12-12=0

所以结果=16+4=20

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