81. Search in Rotated Sorted Array II【leetcode解题报告】

Description

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?

思路

有重复数字之后，则不能单从子数组的左右两端点的大小比较上判断顺序数组所在part。

二分法。

1. 比较A[mid]与target是否相等
   
   • A[mid]==target：返回mid true
   • A[mid]>target or mid

代码

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int n=nums.size();

        int left=0,right=n-1;
        while(left<=right){
            int mid=(left+right)/2;
            if(nums[mid] == target){
                return true;
            }
            if(nums[mid]>target){
                if(mid!=right && nums[mid]==nums[right]){
                    --right;
                }
                else if(nums[mid]>nums[right] && nums[right]>=target){
                    left = mid+1;
                }else{
                    right = mid-1;
                }
            }else{
                if(mid!=left && nums[mid]==nums[left]){
                    ++left;
                }
                else if(nums[mid]<nums[left] && nums[left]<=target){
                    right=mid-1;
                }else{
                    left = mid+1;
                }
            }
        }
        return false;
    }
};

时间复杂度

在没有重复元素时，时间复杂度为$O(lng)$

存在重复元素后，可能存在不停的++left或—right，从而导致复杂度退化到$O(n)$