[Leetcode]Find Minimum in Rotated Sorted Array

Suppose a sorted array 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.

找到最小的元素~和Search in Rotated Array类似，通过边界值和中间值的比较来确定哪半边是升序的，如果左半边是有序的，那么num[left]就是左半边最小的，可以和minVal相比取较小的，然后left = mid + 1; 否则，那么num[mid]是右半边最小的元素，和minVal相比取较小的，然后right = mid - 1；算法复杂度是O(logn)

class Solution:    # @param num, a list of integer    # @return an integer    def findMin(self, num):        if num is None or len(num) == 0: return None        left, right = 0, len(num) - 1        minVal = num[0]        while left <= right:            mid = left + (right - left) / 2            if num[left] <= num[mid]:                minVal = min(minVal, num[left])                left = mid + 1            else:                minVal = min(minVal, num[mid])                right = mid - 1        return minVal

还有一种写法，代码如下~

class Solution:    # @param num, a list of integer    # @return an integer    def findMin(self, num):        if num is None or len(num) == 0: return None        left, right = 0, len(num) - 1        while left < right:            mid = left + (right - left) / 2            if num[right] < num[mid]:                left = mid + 1            else:                right = mid        return num[left]