add 80 solution explain

leetcode/0080.Remove-Duplicates-from-Sorted-Array-II/80. Remove Duplicates from Sorted Array II.go

@@ -2,8 +2,8 @@ package leetcode
 
 func removeDuplicates(nums []int) int {
 	slow := 0
-	for i, v := range nums {
-		if i < 2 || nums[slow-2] != v {
+	for fast, v := range nums {
+		if fast < 2 || nums[slow-2] != v {
 			nums[slow] = v
 			slow++
 		}

leetcode/0080.Remove-Duplicates-from-Sorted-Array-II/README.md

@@ -51,6 +51,6 @@ for (int i = 0; i < len; i++) {
 
 ## 解题思路
 
-这道题和第 26 题很像。是第 26 题的加强版。这道题和第 283 题，第 27 题基本一致，283 题是删除 0，27 题是删除指定元素，这一题是删除重复元素，实质是一样的。
-
-这里数组的删除并不是真的删除，只是将删除的元素移动到数组后面的空间内，然后返回数组实际剩余的元素个数，OJ 最终判断题目的时候会读取数组剩余个数的元素进行输出。
+问题提示有序数组，一般最容易想到使用双指针的解法，双指针的关键点：移动两个指针的条件。
+在该题中移动的条件：快指针从头遍历数组，慢指针指向修改后的数组的末端，当慢指针指向倒数第二个数与快指针指向的数不相等时，才移动慢指针，同时赋值慢指针
+处理边界条件：当数组小于两个元素时，不做处理

website/content/ChapterFour/0001~0099/0080.Remove-Duplicates-from-Sorted-Array-II.md

@@ -2,9 +2,11 @@
 
 ## 题目
 
-Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new length.
+Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new
+length.
 
-Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
+Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra
+memory.
 
 **Example 1**:
 
@@ -34,7 +36,8 @@ It doesn't matter what values are set beyond the returned length.
 
 Confused why the returned value is an integer but your answer is an array?
 
-Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.
+Note that the input array is passed in by reference, which means modification to the input array will be known to the
+caller as well.
 
 Internally you can think of this:
 
@@ -69,10 +72,10 @@ package leetcode
 
 func removeDuplicates(nums []int) int {
 	slow := 0
-	for i, v := range nums {
-		if i < 2 || nums[slow-2] != v{
+	for fast, v := range nums {
+		if fast < 2 || nums[slow-2] != v {
 			nums[slow] = v
-			slow ++
+			slow++
 		}
 	}
 	return slow
@@ -81,7 +84,6 @@ func removeDuplicates(nums []int) int {
 
 ```
 
-
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