旋转数组最小值[有/无重复元素](二分搜索, 注意边界确定)

Author: cls1991

BinarySearch/153_FindMinimumInRotatedSortedArray.py

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+# coding: utf8
+
+
+"""
+    题目链接: https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/description.
+    题目描述:
+
+    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.
+
+"""
+
+
+class Solution(object):
+    def findMin(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+        # return self.binary_search_recursive(nums, 0, len(nums) - 1)
+        return self.binary_search_iterative(nums, 0, len(nums) - 1)
+
+    def binary_search_recursive(self, nums, l, r):
+        if l > r:
+            return nums[r]
+
+        mid = l + (r - l) // 2
+        if nums[mid] < nums[r]:
+            return min(nums[mid], self.binary_search_recursive(nums, l, mid - 1))
+
+        return min(nums[l], self.binary_search_recursive(nums, mid + 1, r))
+
+    def binary_search_iterative(self, nums, l, r):
+        while l <= r:
+            mid = l + (r - l) // 2
+            if nums[mid] < nums[r]:
+                r = mid
+            else:
+                l = mid + 1
+
+        return nums[r]

BinarySearch/154_FindMinimumInRotatedSortedArrayII.py

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+# coding: utf8
+
+
+"""
+    题目链接: https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/description.
+    题目描述:
+
+    Follow up for "Find Minimum in Rotated Sorted Array":
+    What if duplicates are allowed?
+
+    Would this affect the run-time complexity? How and why?
+
+    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.
+
+    The array may contain duplicates.
+
+"""
+
+
+class Solution(object):
+    def findMin(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+        left, right = 0, len(nums) - 1
+        while left <= right:
+            mid = left + (right - left) // 2
+            if nums[mid] < nums[right]:
+                right = mid
+            elif nums[mid] > nums[right]:
+                left = mid + 1
+            else:
+                right -= 1
+
+        return nums[right + 1]