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Commit b5f5e6a

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Merge pull request #51 from Mrzhudky/work
Add Solution 029[Java]
2 parents e8fa938 + 7b619be commit b5f5e6a

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## 两数相除
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### 题目描述
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给定两个整数,被除数 dividend 和除数 divisor。将两数相除,要求不使用乘法、除法和 mod 运算符。
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返回被除数 dividend 除以除数 divisor 得到的商。
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示例 1:
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输入: dividend = 10, divisor = 3
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输出: 3
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示例 2:
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输入: dividend = 7, divisor = -3
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输出: -2
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说明:
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被除数和除数均为 32 位有符号整数。
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除数不为 0。
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假设我们的环境只能存储 32 位有符号整数,其数值范围是 [−2^31 , 2^31 − 1]。
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本题中,如果除法结果溢出,则返回 2^31 − 1。
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### 解法
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1. 考虑用位运算来代替乘除,用二进制表示商,则只要确定了每一个二进制位,则把这些位加和即可得到商;
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2. 对除数进行移位,找到最高位,然后从高到低依次比较每一位对应的数与除数的乘积,若大于则说明商的该位为1,否则为0;
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```java
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class Solution {
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public int divide(int dividend, int divisor) {
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if(dividend == 0 || divisor == 1) {
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return dividend;
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}
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if(divisor == 0 || (dividend == Integer.MIN_VALUE && divisor == -1)) {
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return Integer.MAX_VALUE;
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}
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// 商的符号,true 为正,false 为负
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boolean flag = true;
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if((dividend < 0 && divisor > 0) || (dividend > 0 && divisor < 0)) {
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flag = false;
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}
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long dividendLong = Math.abs((long)dividend);
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long divisorLong = Math.abs((long)divisor);
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int re = 0;
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long factor = 0x1;
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while (dividendLong >= (divisorLong << 1)) {
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divisorLong <<= 1;
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factor <<= 1;
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}
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while (factor > 0 && dividendLong > 0) {
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if(dividendLong >= divisorLong) {
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dividendLong -= divisorLong;
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re += factor;
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}
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factor >>>= 1;
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divisorLong >>>= 1;
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}
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return flag ? re : -re;
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}
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}
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```
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class Solution {
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public int divide(int dividend, int divisor) {
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if(dividend == 0 || divisor == 1) {
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return dividend;
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}
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if(divisor == 0 || (dividend == Integer.MIN_VALUE && divisor == -1)) {
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return Integer.MAX_VALUE;
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}
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// 商的符号,true 为正,false 为负
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boolean flag = true;
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if((dividend < 0 && divisor > 0) || (dividend > 0 && divisor < 0)) {
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flag = false;
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}
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long dividendLong = Math.abs((long)dividend);
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long divisorLong = Math.abs((long)divisor);
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int re = 0;
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long factor = 0x1;
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while (dividendLong >= (divisorLong << 1)) {
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divisorLong <<= 1;
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factor <<= 1;
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}
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while (factor > 0 && dividendLong > 0) {
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if(dividendLong >= divisorLong) {
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dividendLong -= divisorLong;
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re += factor;
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}
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factor >>>= 1;
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divisorLong >>>= 1;
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}
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return flag ? re : -re;
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}
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}

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