154.Find Minimum in Rotated Sorted Array II

Tags: Hard Array Binary Search

Company: Amazon-2, FaceBook-2

Year: 半年内

Links: https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/

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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.

Example 1:

Input: [1,3,5]
Output: 1

Example 2:

Input: [2,2,2,0,1]
Output: 0

Note:

• This is a follow up problem to Find Minimum in Rotated Sorted Array.
• Would allow duplicates affect the run-time complexity? How and why?

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class Solution {
public:
    int findMin(vector<int>& nums) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        int left = 0, right = nums.size() - 1;
        while (left < right) {
            int mid = left + ((right - left) >> 1);
            if (nums[mid] == nums[right]) --right;
            else if (nums[mid] > nums[right]) left = mid + 1;
            else right = mid;
        }
        return nums[right];
    }
};

当数组中存在大量的重复数字时，就会破坏二分查找法的机制，将无法取得 O(lgn) 的时间复杂度，又将会回到简单粗暴的 O(n)，比如这两种情况：{2, 2, 2, 2, 2, 2, 2, 2, 0, 1, 1, 2} 和 {2, 2, 2, 0, 2, 2, 2, 2, 2, 2, 2, 2}，可以发现，当第一个数字和最后一个数字，还有中间那个数字全部相等的时候，二分查找法就崩溃了，因为它无法判断到底该去左半边还是右半边。这种情况下，将右指针左移一位（或者将左指针右移一位），略过一个相同数字，这对结果不会产生影响，因为只是去掉了一个相同的，然后对剩余的部分继续用二分查找法，在最坏的情况下，比如数组所有元素都相同，时间复杂度会升到 O(n)