29. Divide Two Integers

Author: coderbean

src/leetcode/_29_/Main.java

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+package leetcode._29_;
+
+/**
+ * Created by zhangbo54 on 2019-03-04.
+ */
+public class Main {
+    public static void main(String[] args) {
+        Solution solution = new Solution();
+        System.out.println(solution.divide(-2147483648, -1));
+    }
+}
+

src/leetcode/_29_/Solution.java

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+package leetcode._29_;
+
+class Solution {
+    public int divide(int dividend, int divisor) {
+        boolean positive = false;
+        if ((dividend > 0 && divisor > 0) || (dividend < 0 && divisor < 0)) {
+            positive = true;
+        }
+        long longDividend = Math.abs((long) dividend);
+        long longDivisor = Math.abs((long) divisor);
+
+        long result = 0;
+        while (longDividend >= longDivisor) { // 外层循环用来将剩余的继续算
+            long temp = longDivisor;
+            long tempResult = 1;
+            while (longDividend >= temp) { // 内层循环用来加快速度
+                longDividend -= temp;
+                result += tempResult;
+                tempResult <<= 1;
+                temp <<= 1;
+            }
+        }
+        if (positive && result > Integer.MAX_VALUE) {
+            return Integer.MAX_VALUE;
+        }
+        if (!positive && result < Integer.MIN_VALUE) {
+            return Integer.MIN_VALUE;
+        }
+        return positive ? (int) result : (int) -result;
+    }
+}

src/leetcode/_29_/solution.md

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+### [29\. Divide Two Integers](https://leetcode.com/problems/divide-two-integers/)
+
+Difficulty: **Medium**
+
+
+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: [−2<sup>31</sup>,  2<sup>31</sup> − 1]. For the purpose of this problem, assume that your function returns 2<sup>31</sup> − 1 when the division result overflows.
+
+
+#### Solution
+
+Language: **Java**
+
+```java
+class Solution {
+    public int divide(int dividend, int divisor) {
+        boolean positive = false;
+        if ((dividend > 0 && divisor > 0) || (dividend < 0 && divisor < 0)) {
+            positive = true;
+        }
+        long longDividend = Math.abs((long) dividend);
+        long longDivisor = Math.abs((long) divisor);
+​
+        long result = 0;
+        while (longDividend >= longDivisor) { // 外层循环用来将剩余的继续算
+            long temp = longDivisor;
+            long tempResult = 1;
+            while (longDividend >= temp) { // 内层循环用来加快速度
+                longDividend -= temp;
+                result += tempResult;
+                tempResult <<= 1;
+                temp <<= 1;
+            }
+        }
+        if (positive && result > Integer.MAX_VALUE) {
+            return Integer.MAX_VALUE;
+        }
+        if (!positive && result < Integer.MIN_VALUE) {
+            return Integer.MIN_VALUE;
+        }
+        return positive ? (int) result : (int) -result;
+    }
+}
+```