Find Minimum in Rotated Sorted Array(lintcode 159)

Description

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.

Notice:
You may assume no duplicate exists in the array.

Example

Given [4, 5, 6, 7, 0, 1, 2] return 0

Interface

public class Solution {
    /**
     * @param nums: a rotated sorted array
     * @return: the minimum number in the array
     */
    public int findMin(int[] nums) {
        // write your code here

    }
}

Solution

public class Solution {
    /**
     * @param nums: a rotated sorted array
     * @return: the minimum number in the array
     */
    public int findMin(int[] nums) {
        // write your code here
        if (nums == null || nums.length == 0) {
            return -1;
        }

        int start = 0;
        int end = nums.length - 1;

        while (start + 1 < end) {
            int mid = start + (end - start) / 2;

            if (nums[mid] < nums[end]) {
                end = mid;
            } else {
                start = mid;
            }
        }

        if (nums[start] < nums[end]) {
            return nums[start];
        }

        return nums[end];
    }
}

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