添加 problem 26

Algorithms/26. Remove Duplicates from Sorted Array/26. Remove Duplicates from Sorted Array.go

@@ -0,0 +1,56 @@
+package leetcode
+
+func removeDuplicates(nums []int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	last, finder := 0, 0
+	for last < len(nums)-1 {
+		for nums[finder] == nums[last] {
+			finder++
+			if finder == len(nums) {
+				return last + 1
+			}
+		}
+		nums[last+1] = nums[finder]
+		last += 1
+	}
+	return last + 1
+}
+
+func removeDuplicates_(nums []int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	length := len(nums)
+	lastNum := nums[length-1]
+	i := 0
+	for i = 0; i < length-1; i++ {
+		if nums[i] == lastNum {
+			break
+		}
+		if nums[i+1] == nums[i] {
+			removeElement_(nums, i+1, nums[i])
+			// fmt.Printf("此时 num = %v length = %v\n", nums, length)
+		}
+	}
+	return i + 1
+}
+
+func removeElement_(nums []int, start, val int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	j := start
+	for i := start; i < len(nums); i++ {
+		if nums[i] != val {
+			if i != j {
+				nums[i], nums[j] = nums[j], nums[i]
+				j++
+			} else {
+				j++
+			}
+		}
+	}
+	return j
+}

Algorithms/26. Remove Duplicates from Sorted Array/26. Remove Duplicates from Sorted Array_test.go

@@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question26 struct {
+	para26
+	ans26
+}
+
+// para 是参数
+// one 代表第一个参数
+type para26 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans26 struct {
+	one int
+}
+
+func Test_Problem26(t *testing.T) {
+
+	qs := []question26{
+
+		question26{
+			para26{[]int{1, 1, 2}},
+			ans26{2},
+		},
+
+		question26{
+			para26{[]int{0, 0, 1, 1, 1, 1, 2, 3, 4, 4}},
+			ans26{5},
+		},
+
+		question26{
+			para26{[]int{0, 0, 0, 0, 0}},
+			ans26{1},
+		},
+
+		question26{
+			para26{[]int{1}},
+			ans26{1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 26------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans26, q.para26
+		fmt.Printf("【input】:%v    【output】:%v\n", p.one, removeDuplicates(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}

Algorithms/26. Remove Duplicates from Sorted Array/README.md

@@ -0,0 +1,52 @@
+# [26. Remove Duplicates from Sorted Array](https://leetcode.com/problems/remove-duplicates-from-sorted-array/)
+
+## 题目
+
+Given a sorted array nums, remove the duplicates in-place such that each element appear only once 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.
+
+Example 1:
+
+```c
+Given nums = [1,1,2],
+
+Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.
+
+It doesn't matter what you leave beyond the returned length.
+```
+
+Example 2:
+
+```c
+Given nums = [0,0,1,1,1,2,2,3,3,4],
+
+Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.
+
+It doesn't matter what values are set beyond the returned length.
+```
+
+Clarification:
+
+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.
+
+Internally you can think of this:
+
+```c
+// nums is passed in by reference. (i.e., without making a copy)
+int len = removeElement(nums, val);
+
+// any modification to nums in your function would be known by the caller.
+// using the length returned by your function, it prints the first len elements.
+for (int i = 0; i < len; i++) {
+    print(nums[i]);
+}
+```
+
+## 题目大意
+
+这道题和第 27 题很像。给定一个有序数组 nums，对数组中的元素进行去重，使得原数组中的每个元素只有一个。最后返回去重以后数组的长度值。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。