Add solution 153

Author: yanglbme

README.md

@@ -35,6 +35,7 @@ Complete solutions to Leetcode problems, updated daily.
 | 062 | [Unique Paths](https://github.com/yanglbme/leetcode/tree/master/solution/062.Unique%20Paths) | `Array`, `Dynamic Programming` |
 | 063 | [Unique Paths II](https://github.com/yanglbme/leetcode/tree/master/solution/063.Unique%20Paths%20II) | `Array`, `Dynamic Programming` |
 | 082 | [Remove Duplicates from Sorted List II](https://github.com/yanglbme/leetcode/tree/master/solution/082.Remove%20Duplicates%20from%20Sorted%20List%20II) | `Linked List` |
+| 153 | [Find Minimum in Rotated Sorted Array](https://github.com/yanglbme/leetcode/tree/master/solution/153.Find%20Minimum%20in%20Rotated%20Sorted%20Array) | `Array`, `Binary Search` |
 
 
 ### Hard

solution/153.Find Minimum in Rotated Sorted Array/README.md

@@ -0,0 +1,59 @@
+## 寻找旋转排序数组中的最小值
+### 题目描述
+
+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。
+
+示例 1:
+```
+输入: [3,4,5,1,2]
+输出: 1
+```
+
+示例 2:
+```
+输入: [4,5,6,7,0,1,2]
+输出: 0   
+```
+
+### 解法
+利用两个指针 p,q 指示数组的首尾两端，若 nums[p] <= nums[q]，说明数组呈递增趋势，返回 nums[p]；否则判断 nums[p] 与 nums[mid] 的大小，从而决定新的数组首尾两端，循环二分查找。
+
+```java
+class Solution {
+    public int findMin(int[] nums) {
+        int n = nums.length;
+        if (n == 1) {
+            return nums[0];
+        }
+        
+        int p = 0;
+        int q = n - 1;
+        
+        
+        int mid = p + ((q - p) >> 1);
+        
+        while (p < q) {
+            if (nums[p] <= nums[q]) {
+                break;
+            }
+            
+            if (nums[p] > nums[mid]) {
+                q = mid;
+            } else {
+                p = mid + 1;
+            }
+            
+            mid = p + ((q - p) >> 1);
+        }
+        
+        return nums[p];
+        
+    }
+}
+```

solution/153.Find Minimum in Rotated Sorted Array/Solution.java

@@ -0,0 +1,30 @@
+class Solution {
+    public int findMin(int[] nums) {
+        int n = nums.length;
+        if (n == 1) {
+            return nums[0];
+        }
+
+        int p = 0;
+        int q = n - 1;
+
+        int mid = p + ((q - p) >> 1);
+
+        while (p < q) {
+            if (nums[p] <= nums[q]) {
+                break;
+            }
+
+            if (nums[p] > nums[mid]) {
+                q = mid;
+            } else {
+                p = mid + 1;
+            }
+
+            mid = p + ((q - p) >> 1);
+        }
+
+        return nums[p];
+
+    }
+}