diff --git a/Algorithms/0153. Find Minimum in Rotated Sorted Array/153. Find Minimum in Rotated Sorted Array.go b/Algorithms/0153. Find Minimum in Rotated Sorted Array/153. Find Minimum in Rotated Sorted Array.go
new file mode 100644
index 00000000..0c8622d8
--- /dev/null
+++ b/Algorithms/0153. Find Minimum in Rotated Sorted Array/153. Find Minimum in Rotated Sorted Array.go	
@@ -0,0 +1,65 @@
+package leetcode
+
+// 解法一 二分
+func findMin(nums []int) int {
+	low, high := 0, len(nums)-1
+	for low < high {
+		if nums[low] < nums[high] {
+			return nums[low]
+		}
+		mid := low + (high-low)>>1
+		if nums[mid] >= nums[low] {
+			low = mid + 1
+		} else {
+			high = mid
+		}
+	}
+	return nums[low]
+}
+
+// 解法二 二分
+func findMin1(nums []int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	if len(nums) == 1 {
+		return nums[0]
+	}
+	if nums[len(nums)-1] > nums[0] {
+		return nums[0]
+	}
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + (high-low)>>1
+		if nums[low] < nums[high] {
+			return nums[low]
+		}
+		if (mid == len(nums)-1 && nums[mid-1] > nums[mid]) || (mid < len(nums)-1 && mid > 0 && nums[mid-1] > nums[mid] && nums[mid] < nums[mid+1]) {
+			return nums[mid]
+		}
+		if nums[mid] > nums[low] && nums[low] > nums[high] { // mid 在数值大的一部分区间里
+			low = mid + 1
+		} else if nums[mid] < nums[low] && nums[low] > nums[high] { // mid 在数值小的一部分区间里
+			high = mid - 1
+		} else {
+			if nums[low] == nums[mid] {
+				low++
+			}
+			if nums[high] == nums[mid] {
+				high--
+			}
+		}
+	}
+	return -1
+}
+
+// 解法三 暴力
+func findMin2(nums []int) int {
+	min := nums[0]
+	for _, num := range nums[1:] {
+		if min > num {
+			min = num
+		}
+	}
+	return min
+}
diff --git a/Algorithms/0153. Find Minimum in Rotated Sorted Array/153. Find Minimum in Rotated Sorted Array_test.go b/Algorithms/0153. Find Minimum in Rotated Sorted Array/153. Find Minimum in Rotated Sorted Array_test.go
new file mode 100644
index 00000000..43d3cde8
--- /dev/null
+++ b/Algorithms/0153. Find Minimum in Rotated Sorted Array/153. Find Minimum in Rotated Sorted Array_test.go	
@@ -0,0 +1,77 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question153 struct {
+	para153
+	ans153
+}
+
+// para 是参数
+// one 代表第一个参数
+type para153 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans153 struct {
+	one int
+}
+
+func Test_Problem153(t *testing.T) {
+
+	qs := []question153{
+
+		question153{
+			para153{[]int{5, 1, 2, 3, 4}},
+			ans153{1},
+		},
+
+		question153{
+			para153{[]int{1}},
+			ans153{1},
+		},
+
+		question153{
+			para153{[]int{1, 2}},
+			ans153{1},
+		},
+
+		question153{
+			para153{[]int{2, 1}},
+			ans153{1},
+		},
+
+		question153{
+			para153{[]int{2, 3, 1}},
+			ans153{1},
+		},
+
+		question153{
+			para153{[]int{1, 2, 3}},
+			ans153{1},
+		},
+
+		question153{
+			para153{[]int{3, 4, 5, 1, 2}},
+			ans153{1},
+		},
+
+		question153{
+			para153{[]int{4, 5, 6, 7, 0, 1, 2}},
+			ans153{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 153------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans153, q.para153
+		fmt.Printf("【input】:%v    【output】:%v\n", p, findMin(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0153. Find Minimum in Rotated Sorted Array/README.md b/Algorithms/0153. Find Minimum in Rotated Sorted Array/README.md
new file mode 100755
index 00000000..17cf0ab9
--- /dev/null
+++ b/Algorithms/0153. Find Minimum in Rotated Sorted Array/README.md	
@@ -0,0 +1,35 @@
+# [153. Find Minimum in Rotated Sorted Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/)
+
+
+## 题目:
+
+Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
+
+(i.e., `[0,1,2,4,5,6,7]` might become `[4,5,6,7,0,1,2]`).
+
+Find the minimum element.
+
+You may assume no duplicate exists in the array.
+
+**Example 1:**
+
+    Input: [3,4,5,1,2] 
+    Output: 1
+
+**Example 2:**
+
+    Input: [4,5,6,7,0,1,2]
+    Output: 0
+
+
+## 题目大意
+
+假设按照升序排序的数组在预先未知的某个点上进行了旋转。( 例如，数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。请找出其中最小的元素。
+
+你可以假设数组中不存在重复元素。
+
+
+## 解题思路
+
+- 给出一个原本从小到大排序过的数组，但是在某一个分割点上，把数组切分后的两部分对调位置，数值偏大的放到了数组的前部。求这个数组中最小的元素。
+- 求数组最小的元素其实就是找分割点，前一个数比当前数大，后一个数比当前数也要大。可以用二分搜索查找，需要查找的两个有序区间。时间复杂度  O(log n)。这一题也可以用暴力解法，从头开始遍历，动态维护一个最小值即可，时间复杂度 O(n)。