diff --git a/Algorithms/0034. Find First and Last Position of Element in Sorted Array/34. Find First and Last Position of Element in Sorted Array.go b/Algorithms/0034. Find First and Last Position of Element in Sorted Array/34. Find First and Last Position of Element in Sorted Array.go
new file mode 100644
index 00000000..6c678071
--- /dev/null
+++ b/Algorithms/0034. Find First and Last Position of Element in Sorted Array/34. Find First and Last Position of Element in Sorted Array.go	
@@ -0,0 +1,78 @@
+package leetcode
+
+func searchRange(nums []int, target int) []int {
+	return []int{searchFirstEqualElement(nums, target), searchLastEqualElement(nums, target)}
+
+}
+
+// 二分查找第一个与 target 相等的元素，时间复杂度 O(logn)
+func searchFirstEqualElement(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		if nums[mid] > target {
+			high = mid - 1
+		} else if nums[mid] < target {
+			low = mid + 1
+		} else {
+			if (mid == 0) || (nums[mid-1] != target) { // 找到第一个与 target 相等的元素
+				return mid
+			}
+			high = mid - 1
+		}
+	}
+	return -1
+}
+
+// 二分查找最后一个与 target 相等的元素，时间复杂度 O(logn)
+func searchLastEqualElement(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		if nums[mid] > target {
+			high = mid - 1
+		} else if nums[mid] < target {
+			low = mid + 1
+		} else {
+			if (mid == len(nums)-1) || (nums[mid+1] != target) { // 找到最后一个与 target 相等的元素
+				return mid
+			}
+			low = mid + 1
+		}
+	}
+	return -1
+}
+
+// 二分查找第一个大于等于 target 的元素，时间复杂度 O(logn)
+func searchFirstGreaterElement(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		if nums[mid] >= target {
+			if (mid == 0) || (nums[mid-1] < target) { // 找到第一个大于等于 target 的元素
+				return mid
+			}
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+	}
+	return -1
+}
+
+// 二分查找最后一个小于等于 target 的元素，时间复杂度 O(logn)
+func searchLastLessElement(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		if nums[mid] <= target {
+			if (mid == len(nums)-1) || (nums[mid+1] > target) { // 找到最后一个小于等于 target 的元素
+				return mid
+			}
+			low = mid + 1
+		} else {
+			high = mid - 1
+		}
+	}
+	return -1
+}
diff --git a/Algorithms/0034. Find First and Last Position of Element in Sorted Array/34. Find First and Last Position of Element in Sorted Array_test.go b/Algorithms/0034. Find First and Last Position of Element in Sorted Array/34. Find First and Last Position of Element in Sorted Array_test.go
new file mode 100644
index 00000000..52d408aa
--- /dev/null
+++ b/Algorithms/0034. Find First and Last Position of Element in Sorted Array/34. Find First and Last Position of Element in Sorted Array_test.go	
@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question34 struct {
+	para34
+	ans34
+}
+
+// para 是参数
+// one 代表第一个参数
+type para34 struct {
+	nums   []int
+	target int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans34 struct {
+	one []int
+}
+
+func Test_Problem34(t *testing.T) {
+
+	qs := []question34{
+
+		question34{
+			para34{[]int{5, 7, 7, 8, 8, 10}, 8},
+			ans34{[]int{3, 4}},
+		},
+
+		question34{
+			para34{[]int{5, 7, 7, 8, 8, 10}, 6},
+			ans34{[]int{-1, -1}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 34------------------------\n")
+	for _, q := range qs {
+		_, p := q.ans34, q.para34
+		fmt.Printf("【input】:%v       【output】:%v\n", p, searchRange(p.nums, p.target))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0034. Find First and Last Position of Element in Sorted Array/README.md b/Algorithms/0034. Find First and Last Position of Element in Sorted Array/README.md
new file mode 100755
index 00000000..831b0756
--- /dev/null
+++ b/Algorithms/0034. Find First and Last Position of Element in Sorted Array/README.md	
@@ -0,0 +1,36 @@
+# [34. Find First and Last Position of Element in Sorted Array](https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/)
+
+
+## 题目:
+
+Given an array of integers `nums` sorted in ascending order, find the starting and ending position of a given `target` value.
+
+Your algorithm's runtime complexity must be in the order of *O*(log *n*).
+
+If the target is not found in the array, return `[-1, -1]`.
+
+**Example 1:**
+
+    Input: nums = [5,7,7,8,8,10], target = 8
+    Output: [3,4]
+
+**Example 2:**
+
+    Input: nums = [5,7,7,8,8,10], target = 6
+    Output: [-1,-1]
+
+## 题目大意
+
+给定一个按照升序排列的整数数组 nums，和一个目标值 target。找出给定目标值在数组中的开始位置和结束位置。你的算法时间复杂度必须是 O(log n) 级别。如果数组中不存在目标值，返回 [-1, -1]。
+
+
+## 解题思路
+
+- 给出一个有序数组 `nums` 和一个数 `target`，要求在数组中找到第一个和这个元素相等的元素下标，最后一个和这个元素相等的元素下标。
+- 这一题是经典的二分搜索变种题。二分搜索有 4 大基础变种题：
+    1. 查找第一个值等于给定值的元素
+    2. 查找最后一个值等于给定值的元素
+    3. 查找第一个大于等于给定值的元素
+    4. 查找最后一个小于等于给定值的元素
+
+    这一题的解题思路可以分别利用变种 1 和变种 2 的解法就可以做出此题。或者用一次变种 1 的方法，然后循环往后找到最后一个与给定值相等的元素。不过后者这种方法可能会使时间复杂度下降到 O(n)，因为有可能数组中 n 个元素都和给定元素相同。(4 大基础变种的实现见代码)