154 Find Minimum in Rotated Sorted Array II

154. Find Minimum in Rotated Sorted Array II

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 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.

分析& 代码

就比1多了一点点

public int findMin(int[] nums) {
    if(nums.length == 0) {
        return -1;
    }
    int left = 0;
    int right = nums.length - 1;
    while(left < right) {
        int mid = left + (right - left) / 2;
        if(nums[mid] > nums[right]) {
            left = mid + 1;
        } else if(nums[mid] < nums[right]) {
            right = mid;
        } else {
            right--;
        }
    }
    return nums[left];
}

第13行，当nums[mid] == nums[high]的时候high--