Find Minimum in Rotated Sorted Array II.java

/*
Medium Find Minimum in Rotated Sorted Array II My Submissions

40% Accepted
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.

The array may contain duplicates.

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

Tags Expand 
Binary Search Divide and Conqueri

*/

// version 1: just for loop is enough
public class Solution {
    public int findMin(int[] num) {
        //  这道题目在面试中不会让写完整的程序
        //  只需要知道最坏情况下 [1,1,1....,1] 里有一个0
        //  这种情况使得时间复杂度必须是 O(n)
        //  因此写一个for循环就好了。
        //  如果你觉得，不是每个情况都是最坏情况，你想用二分法解决不是最坏情况的情况，那你就写一个二分吧。
        //  反正面试考的不是你在这个题上会不会用二分法。这个题的考点是你想不想得到最坏情况。
        int min = num[0];
        for (int i = 1; i < num.length; i++) {
            if (num[i] < min)
                min = num[i];
        }
        return min;
    }
}

// version 2: use *fake* binary-search
// When num[mid] == num[hi], we couldn't sure the position of minimum in mid's left or right, 
// so just let upper bound reduce one.
public class Solution {
    /**
     * @param num: a rotated sorted array
     * @return: the minimum number in the array
     */
    public int findMin(int[] nums) {
        if (nums == null || nums.length == 0) {
            return -1;
        }
        
        int start = 0;
        int end = nums.length - 1;
        while (start < end) {
            int mid = start + (end - start) / 2;
            if (nums[mid] == nums[end]) {
                // if mid equals to end, that means it's fine to remove end
                // the smallest element won't be removed
                end--;
            } else if (nums[mid] > nums[end]) {
                start = mid + 1;
            } else {
                end = mid;
            }
        }
        
        return nums[start];
    }
}