feat: add 027

Author: Blankj

README.md

@@ -15,7 +15,8 @@
 |19|[Remove Nth Node From End of List][019]|Linked List, Two Pointers|
 |20|[Valid Parentheses][020]|Stack, String|
 |21|[Merge Two Sorted Lists][021]|Linked List|
-|26|[Remove Duplicates from Sorted Array][026]|Linked List|
+|26|[Remove Duplicates from Sorted Array][026]|Array, Two Pointers|
+|27|[Remove Element][027]|Array, Two Pointers|
 
 
 ## Medium
@@ -46,3 +47,4 @@
 [020]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/020/README.md
 [021]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/021/README.md
 [026]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/026/README.md
+[027]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/027/README.md

note/026/README.md

@@ -16,7 +16,7 @@ Your function should return length = `2`, with the first two elements of *nums*
 
 ## 思路
 
-题意是让你从一个有序的数组中移除重复的元素，并返回之后数组的长度。我的思路是判断长度小于等于1的话直接返回原长度即可，否则的话遍历一遍数组，用一个`tail`变量指向尾部，如果后面的元素和前面的元素不同，就让`tail`变量加一，最后返回tail即可。
+题意是让你从一个有序的数组中移除重复的元素，并返回之后数组的长度。我的思路是判断长度小于等于1的话直接返回原长度即可，否则的话遍历一遍数组，用一个`tail`变量指向尾部，如果后面的元素和前面的元素不同，就让`tail`变量加一，最后返回`tail`即可。
 
 ``` java
 public class Solution {

note/027/README.md

@@ -0,0 +1,46 @@
+# [Remove Element][title]
+
+## Description
+
+Given an array and a value, remove all instances of that value in place and return the new length.
+
+Do not allocate extra space for another array, you must do this in place with constant memory.
+
+The order of elements can be changed. It doesn't matter what you leave beyond the new length.
+
+**Example:**
+
+Given input array *nums* = `[3,2,2,3]`, *val* = `3`
+
+Your function should return length = 2, with the first two elements of *nums* being 2.
+
+**Tags:** Array, Two Pointers
+
+
+## 思路
+
+题意是移除数组中值等于`val`的元素，并返回之后数组的长度，并且题目中指定空间复杂度为O(1)，我的思路是用`tail`标记尾部，遍历该数组时当索引元素不等于`val`时，`tail`加一，尾部指向当前元素，最后返回`tail`即可。
+
+``` java
+public class Solution {
+    public int removeElement(int[] nums, int val) {
+        int tail = 0;
+        for (int i = 0, len = nums.length; i < len; ++i) {
+            if (nums[i] != val) {
+                nums[tail++] = nums[i];
+            }
+        }
+        return tail;
+    }
+}
+```
+
+
+## 结语
+
+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]
+
+
+
+[title]: https://leetcode.com/problems/remove-element
+[ajl]: https://github.com/Blankj/awesome-java-leetcode

project/LeetCode/leetcode/src/main/java/com/blankj/easy/_027/Solution.java

@@ -0,0 +1,31 @@
+package com.blankj.easy._027;
+
+/**
+ * <pre>
+ *     author: Blankj
+ *     blog  : http://blankj.com
+ *     time  : 2017/04/21
+ *     desc  :
+ * </pre>
+ */
+
+public class Solution {
+    public int removeElement(int[] nums, int val) {
+        int tail = 0;
+        for (int i = 0, len = nums.length; i < len; ++i) {
+            if (nums[i] != val) {
+                nums[tail++] = nums[i];
+            }
+        }
+        return tail;
+    }
+
+    public static void main(String[] args) {
+        Solution solution = new Solution();
+        int[] data = new int[]{0, 3, 1, 1, 2, 3, 3, 3};
+        int len = solution.removeElement(data, 3);
+        for (int i = 0; i < len; i++) {
+            System.out.print(data[i] + (i == len - 1 ? "" : ", "));
+        }
+    }
+}