Search in Rotated Sorted Array II

medium 原题链接：https://leetcode.com/problems/search-in-rotated-sorted-array-ii/

Search in Rotated Sorted Array II

原题链接：https://leetcode.com/problems/search-in-rotated-sorted-array-ii/

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.

例子

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

Output: true

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?

解法一

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

解法二

二分查找法

class Solution {
public:
    bool search(vector<int>& nums, int target) {
        int first = 0, last = nums.size();
        while (first != last) {
            const int mid = first + (last - first) / 2; 
            if (nums[mid] == target)
                  return true;
            if (nums[first] < nums[mid]) {
                if (nums[first] <= target && target < nums[mid]) 
                    last = mid;
                else
                    first = mid + 1;
            } else if (nums[first] > nums[mid]) {
                if (nums[mid] < target && target <= nums[last-1])
                    first = mid + 1; else
                        last = mid; 
            } else
                  //skip duplicate one
                  first++;
        }
        return false;
    }
};

分析：T=O(n), S=O(1)

如果nums里用重复的数出现是否会增加时间复杂度？

总结