81. Search in Rotated Sorted Array II

Author: coderbean

markdown/0081. Search in Rotated Sorted Array II.md

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+### [81\. Search in Rotated Sorted Array II](https://leetcode.com/problems/search-in-rotated-sorted-array-ii/)
+
+Difficulty: **Medium**
+
+
+Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
+
+(i.e., `[0,0,1,2,2,5,6]` might become `[2,5,6,0,0,1,2]`).
+
+You are given a target value to search. If found in the array return `true`, otherwise return `false`.
+
+**Example 1:**
+
+```
+Input: nums = [2,5,6,0,0,1,2], target = 0
+Output: true
+```
+
+**Example 2:**
+
+```
+Input: nums = [2,5,6,0,0,1,2], target = 3
+Output: false```
+
+**Follow up:**
+
+*   This is a follow up problem to , where `nums` may contain duplicates.
+*   Would this affect the run-time complexity? How and why?
+
+
+#### Solution
+
+Language: **Java**
+
+```java
+class Solution {
+    public boolean search(int[] nums, int target) {
+        int low = 0;
+        int high = nums.length - 1;
+        int middle;
+        while (low <= high) {
+            middle = (high + low) / 2;
+            if (nums[middle] == target) {
+                return true;
+            }
+            if (nums[low] < nums[middle] || nums[high] < nums[middle]) { // [low,middle]是升序 或者是[middle,high]
+                if (target >= nums[low] && nums[middle] >= target) { // 普通二分查找就能得到结果
+                    high = middle - 1;
+                } else {
+                    low = middle + 1;
+                }
+​
+            } else if (nums[high] > nums[middle] || nums[low] > nums[middle]) { // [low,middle] 必然有对称点
+                if (target <= nums[high] && nums[middle] <= target) { // 普通二分查找就能得到结果
+                    low = middle + 1;
+                } else {
+                    high = middle - 1;
+                }
+​
+            } else { // nums[low] == nums[middle] 无法判断是升序还是有对称点
+                high--;
+            }
+        }
+        return false;
+    }
+}
+```
+![pic](https://raw.githubusercontent.com/PicGoBed/PicBed/master/2019-07-29-fwUbnO.jpg)

src/main/java/leetcode/_81_/Main.java

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+package leetcode._81_;
+
+/**
+ * Created by zhangbo54 on 2019-03-04.
+ */
+public class Main {
+    public static void main(String[] args) {
+        Solution solution = new Solution();
+        int[] nums = {2, 5, 6, 0, 0, 1, 2};
+        System.out.println(solution.search(nums, 2));
+        System.out.println(solution.search(nums, 5));
+        System.out.println(solution.search(nums, 6));
+        System.out.println(solution.search(nums, 0));
+        System.out.println(solution.search(nums, 1));
+        System.out.println(solution.search(nums, 2));
+        int[] nums2 = {2,5,6,0,0,1,2};
+        System.out.println(solution.search(nums2, 3));
+    }
+}
+

src/main/java/leetcode/_81_/Solution.java

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+package leetcode._81_;
+
+import java.util.Arrays;
+
+class Solution {
+    public boolean search(int[] nums, int target) {
+        int low = 0;
+        int high = nums.length - 1;
+        int middle;
+        while (low <= high) {
+            middle = (high + low) / 2;
+            if (nums[middle] == target) {
+                return true;
+            }
+            if (nums[low] < nums[middle] || nums[high] < nums[middle]) { // [low,middle]是升序 或者是[middle,high]
+                if (target >= nums[low] && nums[middle] >= target) { // 普通二分查找就能得到结果
+                    high = middle - 1;
+                } else {
+                    low = middle + 1;
+                }
+
+            } else if (nums[high] > nums[middle] || nums[low] > nums[middle]) { // [low,middle] 必然有对称点
+                if (target <= nums[high] && nums[middle] <= target) { // 普通二分查找就能得到结果
+                    low = middle + 1;
+                } else {
+                    high = middle - 1;
+                }
+
+            } else { // nums[low] == nums[middle] 无法判断是升序还是有对称点
+                high--;
+            }
+        }
+        return false;
+    }
+}

src/main/java/leetcode/_81_/solution.md

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+### [81\. Search in Rotated Sorted Array II](https://leetcode.com/problems/search-in-rotated-sorted-array-ii/)
+
+Difficulty: **Medium**
+
+
+Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
+
+(i.e., `[0,0,1,2,2,5,6]` might become `[2,5,6,0,0,1,2]`).
+
+You are given a target value to search. If found in the array return `true`, otherwise return `false`.
+
+**Example 1:**
+
+```
+Input: nums = [2,5,6,0,0,1,2], target = 0
+Output: true
+```
+
+**Example 2:**
+
+```
+Input: nums = [2,5,6,0,0,1,2], target = 3
+Output: false```
+
+**Follow up:**
+
+*   This is a follow up problem to , where `nums` may contain duplicates.
+*   Would this affect the run-time complexity? How and why?
+
+
+#### Solution
+
+Language: **Java**
+
+```java
+class Solution {
+    public boolean search(int[] nums, int target) {
+        int low = 0;
+        int high = nums.length - 1;
+        int middle;
+        while (low <= high) {
+            middle = (high + low) / 2;
+            if (nums[middle] == target) {
+                return true;
+            }
+            if (nums[low] < nums[middle] || nums[high] < nums[middle]) { // [low,middle]是升序 或者是[middle,high]
+                if (target >= nums[low] && nums[middle] >= target) { // 普通二分查找就能得到结果
+                    high = middle - 1;
+                } else {
+                    low = middle + 1;
+                }
+​
+            } else if (nums[high] > nums[middle] || nums[low] > nums[middle]) { // [low,middle] 必然有对称点
+                if (target <= nums[high] && nums[middle] <= target) { // 普通二分查找就能得到结果
+                    low = middle + 1;
+                } else {
+                    high = middle - 1;
+                }
+​
+            } else { // nums[low] == nums[middle] 无法判断是升序还是有对称点
+                high--;
+            }
+        }
+        return false;
+    }
+}
+```
+![pic](https://raw.githubusercontent.com/PicGoBed/PicBed/master/2019-07-29-fwUbnO.jpg)