20181208 QianHu

QianHu/LeetCode/33-SearchInRotatedSortedArray/__init__.py

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+# -*- coding: utf-8 -*-

QianHu/LeetCode/33-SearchInRotatedSortedArray/main.py

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+# -*- coding: utf-8 -*-
+
+
+class Solution(object):
+    def search(self, nums, target):
+        """
+        :type nums: List[int]
+        :type target: int
+        :rtype: int
+        """
+        low = 0
+        high = len(nums)
+        while low < high:
+            mid = (low + high) // 2
+            if target == nums[mid]:
+                return mid
+            if nums[mid] < nums[low]:
+                if target < nums[mid] or target >= nums[low]:
+                    high = mid
+                else:
+                    low = mid + 1
+            elif nums[mid] > nums[high - 1]:
+                if target > nums[mid] or target < nums[low]:
+                    low = mid + 1
+                else:
+                    high = mid
+            else:
+                if target > nums[mid]:
+                    low = mid + 1
+                else:
+                    high = mid
+        return -1
+
+
+if __name__ == '__main__':
+    solution = Solution()
+    nums = [8, 1, 2, 3, 4, 5, 6, 7]
+    target = 6
+    assert solution.search(nums, target) == 6

QianHu/LeetCode/33-SearchInRotatedSortedArray/readme.md

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+# LeetCode - 33. Search in Rotated Sorted Array
+
+link: [https://leetcode.com/problems/search-in-rotated-sorted-array/](https://leetcode.com/problems/search-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]).
+
+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
+```
+Accepted 334,119  Submissions 1,033,865
+
+
+
+## 题目来源与背景故事
+
+LeetCode上一道Medium题目，2018年1月面试Face++ Python后端开发时遇到过一样的题目，当时回答的不好，一直在纠结C点的位置关系；在参加2018年清华大学研究生入学考试，计算机基础知识综合912中，也遇到了类似的题目，那道题目和此题略有区别，将下图中的BC交换了，实际是一个类似y = 1 - |x|形状，即AC段单调递增，CB段单调递减，这题比较简单，所以回答的还可以
+
+
+
+
+## 问题描述
+
+给定一个有序数组，将其中前面部分，放至后面，在这样的数组中查找特定值，要求复杂度O(log N)
+
+![ex1](img_ex1.jpeg)
+
+## 题目分析
+此题目实际是是一个二分查找的应用，需要对原先二分查找的基础上修改条件判断
+
+我们将low设置为A处，high设置在B处（右侧），mid的位置就很关键了，mid可能会落在AC之间，也可能在CB之间，接下来分情况考虑，更定输入nums，查找目标target
+
+
+
+### 1. mid恰好就是target所在位置
+
+这是最理想的情况，也是循环结束的条件之一
+
+
+
+### 2. mid落在[A, C)之间
+
+![ex2](img_ex2.jpeg)
+
+此时，mid的特点是，nums[C] < nums[B] < nums[low] <= nums[mid] <= nums[C] 
+
+####  2.1 target位于[low, mid)之间 
+
+此时如果target位于[low, mid)，那么target的特点是，nums[low] <= nums[target] < nums[mid]，此时target位于完全有序的[low, mid)区间，退化成了一个标准的二分查找
+
+#### 2.2 target位于(mid, high)之间
+
+此时如果target位于位于(mid, high)之间，此时退化成一个规模更小的旋转数组查找，此时target有两种情况
+
+- 位于(mid, C)之间，此时target的特点是nums[mid] < target && nums[low] < target
+- 位于[C, B)之间，此时target特点是nums[mid] > target && nums[low] > target 
+
+
+
+### 3. mid落在[C, B)之间
+
+![ex3](img_ex3.jpeg)
+
+此时，mid的特点是，nums[C]  <= nums[low] <= nums[B] < nums[mid] < nums[C] 
+
+#### 2.1 target位于[low, mid)之间
+
+此时如果target位于[low, mid)，那么target的特点是， nums[target] < nums[low] < nums[mid]，此时退化成一个规模更小的旋转数组查找，此时target有两种情况
+
+- 位于[low, C)之间，此时target的特点是nums[mid] < target &&  target >= nums[low]
+- 位于[C, mid)之间，此时target特点是nums[mid] > target && nums[low] > target 
+
+#### 2.2 target位于(mid, high)之间
+
+此时如果target位于(mid, high)，那么target的特点是，nums[mid] < nums[target] < nums[high]，此时target位于完全有序的(mid, high)区间，退化成了一个标准的二分查找
+
+
+
+## 代码实现
+
+```python
+# -*- coding: utf-8 -*-
+
+
+class Solution(object):
+    def search(self, nums, target):
+        """
+        :type nums: List[int]
+        :type target: int
+        :rtype: int
+        """
+        low = 0
+        high = len(nums)
+        while low < high:
+            mid = (low + high) // 2
+            if target == nums[mid]:
+                return mid
+            if nums[mid] < nums[low]:
+                if target < nums[mid] or target >= nums[low]:
+                    high = mid
+                else:
+                    low = mid + 1
+            elif nums[mid] > nums[high - 1]:
+                if target > nums[mid] or target < nums[low]:
+                    low = mid + 1
+                else:
+                    high = mid
+            else:
+                if target > nums[mid]:
+                    low = mid + 1
+                else:
+                    high = mid
+        return -1
+
+
+if __name__ == '__main__':
+    solution = Solution()
+    nums = [8, 1, 2, 3, 4, 5, 6, 7]
+    target = 6
+    assert solution.search(nums, target) == 6
+
+```
+
+
+
+## 运行结果
+
+| Time Submitted | Status                                   | Runtime | Language |
+| -------------- | ---------------------------------------- | ------- | -------- |
+| a day ago      | [Accepted](https://leetcode.com/submissions/detail/193848779/) | 24 ms   | python   |

QianHu/LeetCode/__init__.py

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+# -*- coding: utf-8 -*-

QianHu/readme.md

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 # QianHu贡献内容
 
-1.真二叉树重构
+## Tsinghua DS MOOC
+### 1.真二叉树重构（ProperRebuild）
 
+## LeetCode
+### 33.Search in Rotated Sorted Array