Find Minimum in Rotated Sorted Array II

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.

Solution:

If we found duplicate, we need to move start or end. In ideal case, we should set start = mid or end = mid. However, since the array is rotated, the duplicate numbers could be appeared in both side of the array, for exemple: [4,4,5,6,7,0,1,2,4,4]. Therefore, we can only move start or end by one step at a time. For instance: if nums[mid] == nums[start], start++. If nums[mid] == nums[end], end--.

Code:

public int findMin(int[] nums) {
    if (nums == null || nums.length == 0) {
        return Integer.MAX_VALUE;
    }
    int start = 0;
    int end = nums.length - 1;
    int target = nums[end];
    while (start + 1 < end) {
        int mid = start + (end - start) / 2;
        if (nums[start] == nums[mid]) {
            start++;
        } else if (nums[end] == nums[mid]) {
            end--;
        } else if (nums[mid] <= target) {
            end = mid;
        } else {
            start = mid;
        }
    }
    if (nums[start] < nums[end]) {
        return nums[start];
    } else {
        return nums[end];
    }
}