LeetCode算法系列：81. Search in Rotated Sorted Array II

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题目描述：

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?

算法实现：

和LeetCode的第33题很类似33. Search in Rotated Sorted Array，只是因为出现了重复元素，我无法再用二分法找到旋转的轴，取而代之的是扫描的方式

class Solution {
public:
    bool search(vector<int>& nums, int target) {
        if(nums.empty())return false;
        int i,j,k;
        k = 0;
        while(k + 1 < nums.size() && nums[k] <= nums[k + 1])k ++;
        k ++;

        if(target == nums[0])return true;
        else if(target < nums[0]){
            i = k;
            j = nums.size() - 1;
        } 
        else i = 1, j = k - 1;
        while(i <= j){
            k = int((i + j)/2);
            if(nums[k] == target)return true;
            else if(nums[k] > target) j = k - 1;
            else i = k + 1;
        }
        return false;
    }
};