feat: add the solution of Single Number(136) with kotlin.

Author: relish-wang

README.md

@@ -86,6 +86,7 @@
 | [121][121-question] | [Best Time to Buy and Sell Stock][121-tips]                  | [Easy][E]   | [✅][121-java] |              | [✅][121-kotlin] |
 | [122][122-question] | [Best Time to Buy and Sell Stock II][122-tips]               | [Easy][E]   | [✅][122-java] |              | [✅][122-kotlin] |
 | [125][125-question] | [Valid Palindrome][125-tips]                                 | [Easy][E]   |                |              | [✅][125-kotlin] |
+| [136][136-question] | [Single Number][136-tips]                                    | [Easy][E]   |                |              | [✅][136-kotlin] |
 | [226][226-question] | [Invert Binary Tree][226-tips]                               | [Easy][E]   | [✅][226-java] | [✅][226-js] | [✅][226-kotlin] |
 | [504][504-question] | [Base 7][504-tips]                                           | [Easy][E]   |                |              | [✅][504-kotlin] |
 | [543][543-question] | [Diameter of Binary Tree][543-tips]                          | [Easy][E]   | [✅][543-java] |              | [✅][543-kotlin] |
@@ -206,6 +207,7 @@
 [121-question]: https://leetcode.com/problems/best-time-to-buy-and-sell-stock/description/
 [122-question]: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/description/
 [125-question]: https://leetcode.com/problems/valid-palindrome/
+[136-question]: https://leetcode.com/problems/single-number/
 [226-question]: https://leetcode.com/problems/invert-binary-tree/
 [504-question]: https://leetcode.com/problems/base-7/description/
 [543-question]: https://leetcode.com/problems/diameter-of-binary-tree/
@@ -307,6 +309,7 @@
 [121-tips]: ./tips/121/README.md
 [122-tips]: ./tips/122/README.md
 [125-tips]: ./tips/125/README.md
+[136-tips]: ./tips/136/README.md
 [226-tips]: ./tips/226/README.md
 [504-tips]: ./tips/504/README.md
 [543-tips]: ./tips/543/README.md
@@ -529,6 +532,7 @@
 [121-kotlin]: ./src/_121/kotlin/Solution.kt
 [122-kotlin]: ./src/_122/kotlin/Solution.kt
 [125-kotlin]: ./src/_125/kotlin/Solution.kt
+[136-kotlin]: ./src/_136/kotlin/Solution.kt
 [226-kotlin]: ./src/_226/kotlin/Solution.kt
 [504-kotlin]: ./src/_504/kotlin/Solution.kt
 [543-kotlin]: ./src/_543/kotlin/Solution.kt

src/_136/kotlin/Solution.kt

@@ -0,0 +1,38 @@
+package _136.kotlin
+
+import java.util.*
+import kotlin.collections.HashMap
+
+/**
+ * @author relish
+ * @since 2018/11/14
+ */
+class Solution {
+    fun singleNumber1(nums: IntArray): Int {
+        val c = HashSet<Int>()
+        val r = HashSet<Int>()
+        for (num in nums) {
+            if (r.contains(num)) continue
+            if (c.contains(num)) {
+                c.remove(num)
+                r.add(num)
+            } else {
+                c.add(num)
+            }
+        }
+        return c.single()
+    }
+
+    fun singleNumber(nums: IntArray): Int {
+        var s = 0
+        for (num in nums) {
+            s = s.xor(num)
+        }
+        return s
+    }
+}
+
+fun main(args: Array<String>) {
+    println(Solution().singleNumber(intArrayOf(2, 2, 1)))
+    println(Solution().singleNumber(intArrayOf(4, 1, 2, 1, 2)))
+}

tips/136/README.md

@@ -0,0 +1,71 @@
+[Single Number][title]
+
+## Description
+Given a non-empty array of integers, every element appears twice except for one. Find that single one.
+
+**Note:**
+
+Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
+
+**Example 1:**
+
+```
+Input: [2,2,1]
+Output: 1
+```
+
+**Example 1:**
+
+```
+Input: [4,1,2,1,2]
+Output: 4
+```
+
+
+**Tags:** 
+HashTable、 Bit Manipulation
+
+
+## 思路 1
+1 存在`r`内的属于已出现重复的数字
+2 存在`c`内的属于可能只出现一次的数字
+3 当当前数字存在r中时, 说明是重复数字, 直接跳过进入下个循环;当出现重复数字时, 在r中移除该数字, 在c中加入该数字
+4 最后`c`中就只剩下一个数字了
+从代码看上去是线性复杂度的, 其实并不是, 因为contains方法里执行了循环操作。因此此解法效率较低。但可以应对其他数字出现2次以上的情况。
+kotlin(272ms/62.96%): 
+```kotlin
+fun singleNumber(nums: IntArray): Int {
+    val c = HashSet<Int>()
+    val r = HashSet<Int>()
+    for (num in nums) {
+        if (r.contains(num)) continue
+        if (c.contains(num)) {
+            c.remove(num)
+            r.add(num)
+        } else {
+            c.add(num)
+        }
+    }
+    return c.single()
+}
+```
+
+## 思路 2
+异或。(一个数字与自己异或为0;任何数与0异或为这个数本身)
+kotlin(228ms/100.00%): 
+```kotlin
+fun singleNumber(nums: IntArray): Int {
+    var s = 0
+    for (num in nums) {
+        s = s.xor(num)
+    }
+    return s
+}
+```
+
+## 结语
+   
+如果你同我们一样热爱数据结构、算法、LeetCode，可以关注我们 GitHub 上的 LeetCode 题解：[LeetCode-Solution][ls]
+
+[title]: https://leetcode.com/problems/single-number/
+[ls]: https://github.com/RichCodersAndMe/LeetCode-Solution