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‎README.md

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| 19 | [Remove Nth Node From End of List][019] | Linked List, Two Pointers |
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| 22 | [Generate Parentheses][022] | String, Backtracking |
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| 24 | [Swap Nodes in Pairs][024] | Linked List |
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| 29 | [Divide Two Integers][029] | Math, Binary Search |
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| 33 | [Search in Rotated Sorted Array][033] | Arrays, Binary Search |
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| 43 | [Multiply Strings][043] | Math, String |
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| 49 | [Group Anagrams][049] | Hash Table, String |
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[019]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/019/README.md
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[022]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/022/README.md
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[024]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/024/README.md
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[029]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/029/README.md
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[033]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/033/README.md
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[043]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/043/README.md
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[049]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/049/README.md

‎note/029/README.md

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# [Divide Two Integers][title]
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## Description
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Divide two integers without using multiplication, division and mod operator.
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If it is overflow, return MAX_INT.
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**Tags:** Math, Binary Search
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## 思路
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题意是让你算两个整型数相除后的结果,如果结果溢出就返回 `MAX_INT`,但不能使用乘、除、余的操作符,如果是用加减操作符的话肯定会超时哈,这样的话我们就只能想到位操作符了。
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首先,我们分析下溢出的情况,也就是当被除数为 `Integer.MIN_VALUE`,除数为 `-1` 时会溢出,因为 `|Integer.MIN_VALUE| = Integer.MAX_VALUE + 1`。
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然后,我们把除数和被除数都转为 `long` 类型的正数去做下一步操作,我这里以 `22` 和 `3` 相除为例子,因为 `22 >= 3`,我们对 `3` 进行左移一位,也就是乘 2,结果为 `6`,比 `22` 小,我们继续对 `6` 左移一位结果为 `12`,还是比 `22` 小,我们继续对 `12` 左移一位为 `24`,比 `22` 大,这时我们可以分析出,`22` 肯定比 `3` 的 `4` 倍要大,`4` 是怎么来的?因为我们左移了两次,也就是 `1 << 2 = 4`,此时我们记下这个 `4`,然后让 `22 - 3 * 4 = 10`,因为 `10 >= 3`,对 `10` 和 `3` 进行同样的操作,我们可以得到 `2`,此时加上上次的 `4`,和为 `6`,也就是 `22` 比 `3` 的 `6` 倍要大,此时还剩余 `10 - 6 = 4`,因为 `4 >= 3`,所以对 `4` 和 `3` 进行同样的操作,我们发现并不能对 `3` 进行左移了,因为 `4 >= 3`,所以 `1` 倍还是有的,所以加上最后的 `1`,结果为 `6 + 1 = 7`,也就是 `22` 整除 `3` 结果为 `7`,根据以上思路来书写如下代码应该不是难事了吧。
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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 == Integer.MIN_VALUE && divisor == -1) {
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return Integer.MAX_VALUE;
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}
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long dvd = Math.abs((long) dividend);
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long dvr = Math.abs((long) divisor);
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int res = 0;
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while (dvd >= dvr) {
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long temp = dvr, multiple = 1;
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while (dvd >= temp << 1) {
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temp <<= 1;
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multiple <<= 1;
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}
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dvd -= temp;
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res += multiple;
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}
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return (dividend < 0) ^ (divisor < 0) ? -res : res;
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}
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}
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```
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## 结语
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如果你同我一样热爱数据结构、算法、LeetCode,可以关注我 GitHub 上的 LeetCode 题解:[awesome-java-leetcode][ajl]
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[title]: https://leetcode.com/problems/divide-two-integers
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[ajl]: https://github.com/Blankj/awesome-java-leetcode
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package com.blankj.medium._029;
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/**
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* <pre>
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* author: Blankj
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* blog : http://blankj.com
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* time : 2018年01月31日
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* desc :
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* </pre>
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*/
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public class Solution {
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public int divide(int dividend, int divisor) {
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if (dividend == Integer.MIN_VALUE && divisor == -1) {
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return Integer.MAX_VALUE;
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}
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long dvd = Math.abs((long) dividend);
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long dvr = Math.abs((long) divisor);
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int res = 0;
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while (dvd >= dvr) {
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long temp = dvr, multiple = 1;
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while (dvd >= temp << 1) {
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temp <<= 1;
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multiple <<= 1;
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}
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dvd -= temp;
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res += multiple;
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}
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return (dividend < 0) ^ (divisor < 0) ? -res : res;
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}
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public static void main(String[] args) {
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Solution solution = new Solution();
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System.out.println(solution.divide(-2147483648, 1));
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}
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}

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