81. Search in Rotated Sorted Array II

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]).

You are given a target value to search. If found in the array return true, otherwise return false.

Example 1:

Input: nums = [2,5,6,0,0,1,2], target = 0
Output: true

Example 2:

Input: nums = [2,5,6,0,0,1,2], target = 3
Output: false

Follow up:

• This is a follow up problem to Search in Rotated Sorted Array, where nums may contain duplicates.
• Would this affect the run-time complexity? How and why?

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这一题在LeetCode Search in Rotated Sorted Array的基础上，增加了数组有重复元素的约束，其实相当于LeetCode Search in Rotated Sorted Array和LeetCode Find Minimum in Rotated Sorted Array II的结合，搜索框架采用前者（即先找到有序的半个部分，再决定丢弃哪半个部分），边界判断采用后者（即遇到中间和边缘相等时，移动边缘）。代码上只是在前者代码的基础上增加了一个else分支。完整代码如下：

class Solution {
public:
    bool search(vector<int>& nums, int target)
    {
        int l = 0, r = nums.size() – 1;
        while (l <= r) {
            int mid = (l + r) / 2;
            if (nums[mid] == target)
                return true;
            if (nums[mid] < nums[r]) {
                if (target > nums[mid] && target <= nums[r]) {
                    l = mid + 1;
                }
                else {
                    r = mid – 1;
                }
            }
            else if (nums[mid] > nums[r]) {
                if (target >= nums[l] && target < nums[mid]) {
                    r = mid – 1;
                }
                else {
                    l = mid + 1;
                }
            }
            else {
                –r;
            }
        }
        return false;
    }
};

本代码提交AC，用时12MS。因为多了一个else分支，算法最坏情况下，会达到O(n)的复杂度。

二刷。思路相同，使用递归实现，代码如下：

class Solution {
public:
	bool SearchRange(vector<int>& nums, int l, int r, int target) {
		if (l > r)return false;
		if (l == r)return nums[l] == target;

		int mid = l + (r - l) / 2;
		if (nums[mid] == target)return true;

		if (nums[l] < nums[mid]){
			if (target >= nums[l] && target < nums[mid])return SearchRange(nums, l, mid - 1, target);
			else return SearchRange(nums, mid + 1, r, target);
		}
		else if (nums[l] > nums[mid]) {
			if (target > nums[mid] && target <= nums[r])return SearchRange(nums, mid + 1, r, target);
			else return SearchRange(nums, l, mid - 1, target);
		}
		else {
			++l;
			return SearchRange(nums, l, r, target);
		}
	}
	bool search(vector<int>& nums, int target) {
		return SearchRange(nums, 0, nums.size() - 1, target);
	}
};

本代码提交AC，用时4MS。