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?

AC代码

class Solution {
public:
    bool search(vector<int>& nums, int target) {
        if (nums.empty()) return false;
        int left = 0, right = nums.size() - 1, mid;
        while (left <= right) {
            mid = (left + right) / 2;
            if (nums[mid] == target) return true;
            if (nums[left]==nums[right]) left++; //或者right--         
            else if (nums[mid] >= nums[left]) {  // mid左边有序
                if (target >= nums[left] && target < nums[mid]) right = mid - 1;
                else left = mid + 1;
            }
            else if (nums[mid] <= nums[right]) {  // mid右边有序
                if (target >= nums[mid] && target <= nums[right]) left = mid + 1;
                else right = mid - 1;
            }
        }
        return false;
    }
};

总结

这个题与33题唯一的不同是左右两端的值可能相同，修改方法是当这种情况发生时，执行一次left++或者right--操作