在旋转数组中找最小的值

原题如下所示：

153. Find Minimum in Rotated Sorted Array

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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.

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public class Solution {
public int findMin(int[] nums) {
if (nums==null || nums.length==0) { return Integer.MIN_VALUE; }
int left = 0, right = nums.length-1;
while (left < right-1) { // while (left < right-1) is a useful technique
int mid = left + (right-left)/2;
if (nums[mid] > nums[right]) { left = mid; }
else { right = mid; }
}
if (nums[left] > nums[right]) { return nums[right]; }
return nums[left];
}
}

如上所示，每次与最右边的元素比较会避免很多特殊的情况，这是由最有元素的特殊性决定的，mid元素大于right说明当前处于较大的组中，当mid小于right元素的值则说明当前处于较小的组。而且有效的避免了完全导致，和完全未倒置的情况。当完全倒置时  每次与最右端比较持续右移，当完全未旋转时 持续左移。