33. Search in Rotated Sorted Array

coderbean · coderbean · commit 913a753c671e · 2019-03-14T20:21:46.000+08:00
diff --git a/src/leetcode/_33_/Main.java b/src/leetcode/_33_/Main.java
@@ -0,0 +1,12 @@
+package leetcode._33_;
+
+/**
+ * Created by zhangbo54 on 2019-03-04.
+ */
+public class Main {
+    public static void main(String[] args) {
+        Solution solution = new Solution();
+        System.out.println(solution.search(new int[]{5, 1, 3}, 3));
+    }
+}
+
diff --git a/src/leetcode/_33_/Solution.java b/src/leetcode/_33_/Solution.java
@@ -0,0 +1,32 @@
+package leetcode._33_;
+
+import java.util.Arrays;
+
+class Solution {
+    public int search(int[] nums, int target) {
+        int low = 0;
+        int high = nums.length - 1;
+        while (low <= high) {
+            int mid = (low + high) / 2;
+            if (nums[mid] == target) {
+                return mid;
+            }
+            if (nums[mid] >= nums[low]) { //说明low到mid这一段是递增的，不会经过对称线
+                if (target >= nums[low] && target <= nums[mid]) { // 如果 target 在 这之间，继续二分即可
+                    int result = Arrays.binarySearch(nums, low, mid, target);
+                    return result >= 0 ? result : -1;
+                } else {
+                    low = mid + 1;
+                }
+            } else {
+                if (target >= nums[mid] && target <= nums[high]) { // 如果 target 在 这之间，继续二分即可
+                    int result = Arrays.binarySearch(nums, mid + 1, high + 1, target);
+                    return result >= 0 ? result : -1;
+                } else {
+                    high = mid - 1;
+                }
+            }
+        }
+        return -1;
+    }
+}
diff --git a/src/leetcode/_33_/solution.md b/src/leetcode/_33_/solution.md
@@ -0,0 +1,64 @@
+### [33\. Search in Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/)
+
+Difficulty: **Medium**
+
+
+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]`).
+
+You are given a target value to search. If found in the array return its index, otherwise return `-1`.
+
+You may assume no duplicate exists in the array.
+
+Your algorithm's runtime complexity must be in the order of _O_(log _n_).
+
+**Example 1:**
+
+```
+Input: nums = [4,5,6,7,0,1,2], target = 0
+Output: 4
+```
+
+**Example 2:**
+
+```
+Input: nums = [4,5,6,7,0,1,2], target = 3
+Output: -1```
+
+
+#### Solution
+
+Language: **Java**
+
+```java
+class Solution {
+    public int search(int[] nums, int target) {
+        int low = 0;
+        int high = nums.length - 1;
+        while (low <= high) {
+            int mid = (low + high) / 2;
+            if (nums[mid] == target) {
+                return mid;
+            }
+            if (nums[mid] >= nums[low]) { //说明low到mid这一段是递增的，不会经过对称线
+                if (target >= nums[low] && target <= nums[mid]) { // 如果 target 在 这之间，继续二分即可
+                    int result = Arrays.binarySearch(nums, low, mid, target);
+                    return result >= 0 ? result : -1;
+                } else {
+                    low = mid + 1;
+                }
+            } else {
+                if (target >= nums[mid] && target <= nums[high]) { // 如果 target 在 这之间，继续二分即可
+                    int result = Arrays.binarySearch(nums, mid + 1, high + 1, target);
+                    return result >= 0 ? result : -1;
+                } else {
+                    high = mid - 1;
+                }
+            }
+        }
+        return -1;
+    }
+}
+```
+![](https://ws1.sinaimg.cn/large/006tKfTcgy1g12lb7o7nfj30u00v5wk7.jpg)
