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.

题意：给出一个旋转后的数组，求该数组中的最小值。

思路：用二分法的递归形式。（1）数组只有一个元素，这个就是最小值 。（2）数组中有两个元素，求这两个的最小值。（3）如果第一个元素小于最后一个元素，说明这个区间的数组是从小到大排序的。最左边的就是最小值。（4）求这个区间的中间位置mid,然后递归计算[low, mid]和[mid+1,high]的最小值 。

代码如下：

class Solution{    private int findMin(int[] nums, int low, int high)    {        if (low == high) return nums[low];        else if (1 == high - low)        {            return Math.min(nums[low], nums[high]);        }        else        {            if (nums[low] <  nums[high]) return nums[low];            int mid = (low + high) >> 1;            int min1 = findMin(nums, low, mid);            int min2 = findMin(nums, mid + 1, high);            return Math.min(min1, min2);        }    }    public int findMin(int[] nums)    {        int high = nums.length - 1;        return findMin(nums, 0, high);    }}

解法二：

class Solution {    public int findMin(int[] nums)    {        int left = 0, right = nums.length - 1;        while (left < right)        {            int mid = (left + right) >> 1;            if (nums[mid] < nums[right])            {                right = mid;            }            else if (nums[mid] > nums[right])            {                left = mid + 1;            }        }        return nums[left];    }}