154. Find Minimum in Rotated Sorted Array II

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Follow up for "Find Minimum in Rotated Sorted Array":
What if duplicates are allowed?
Would this affect the run-time complexity? How and why?

Suppose an array sorted in ascending order 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.

解法1：

当有重复数字的时候，最差的情况会退化成O(N)的算法。在原来算法的基础上还要判断两个数字是否相等。如果相等的时候，只能排除掉一个数而不是一半的数字。
C++

1

Java

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