Leetcode 153. Find Minimum in Rotated Sorted Array

Description: Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:

[4,5,6,7,0,1,2] if it was rotated 4 times.
[0,1,2,4,5,6,7] if it was rotated 7 times.

Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].

Given the sorted rotated array nums of unique elements, return the minimum element of this array.

Link: 153. Find Minimum in Rotated Sorted Array

Examples:

Example 1:
Input: nums = [3,4,5,1,2]
Output: 1
Explanation: The original array was [1,2,3,4,5] rotated 3 times.

Example 2:
Input: nums = [4,5,6,7,0,1,2]
Output: 0
Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.

Example 3:
Input: nums = [11,13,15,17]
Output: 11
Explanation: The original array was [11,13,15,17] and it was rotated 4 times.

思路: 找到被旋转的数组中的最小值，给定的target值有确定的值，可以直接找，那么最少值呢？它在这个数组中有什么特点，可以作为一个标准找到呢？如果一个数的前面的值大于它，它就是最小的，如果一个值后面的值小于它，那么它后面的值就是最小值。其他情况就是有序的情况，移动指针就可以。注意为了使mid+1不越界，设置 while l < r,不是以前的 while l <= r。

class Solution(object):
    def findMin(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        n = len(nums)
        l, r = 0, n-1
        while l < r:
            mid = int((l+r)/2)
            if nums[mid-1] > nums[mid]:
                return nums[mid]
            elif nums[mid+1] < nums[mid]:
                return nums[mid+1]
            if nums[l] < nums[mid]:
                l = mid+1    
            else:
                r = mid-1
        return nums[0]

这样做虽然是正确的，却没有通透的理解题目，利用旋转数组的旋转这一信息。数组被旋转，也就是被从被从某处截断，将前面的一段移接到后面，那么如果数组被旋转过，其右边的部分是最小值所在的区间，右边的部分要小于左边的部分。那么我们可以通过mid与right的数值比较确定mid所在的位置。如果nums[mid] < nums[right]，说明mid已经在被截的这后面一部分了，有可能是最小值的位置，也可能是最小值后面的位置，right = mid, 否则如果nums[mid] > nums[right]，说明mid在前面那一部分，也就是数值大的那一部分，所以l = mid+1. 直到区间缩小到 l==r,就是最小值。

class Solution(object):
    def findMin(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        n = len(nums)
        l, r = 0, n-1
        while l < r:
            mid = int((l+r)/2)
            if nums[mid] < nums[r]:
                r = mid    
            else:
                l = mid+1
        return nums[l]

这样才更清晰，更易懂，要明白旋转后的数组，最小值在后半段，最大值在前半段。

日期: 2021-04-11