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.

分析:

排好序的数组基于某个位置截断然后前后两段交换一下，

[0,1,2], [4,5,6,7] => [4,5,6,7], [0,1,2]

前后两段都应该是递增序列，所以遍历列表，如果某个元素小于前一个，那就是截断的第一段序列的开始，就是要找的整个列表里面的最小值。

class Solution:

    # @param num, a list of integer
    # @return an integer
    def findMin(self, num):
        pre = num[0]
        for i in num[1:]:
            if i < pre:
                return i
            else:
                pre = i
        return num[0]

if __name__ == ‘__main__‘:
    s = Solution()
    assert s.findMin([4, 5, 6, 7, 0, 1, 2]) == 0
    print ‘PASS‘

小结：

这个问题从直觉上就是这个解法，直接写了代码就Accepted了。

Find Minimum in Rotated Sorted Array

原文：http://www.cnblogs.com/openqt/p/4033553.html