LeetCode-81-Search in Rotated Sorted Array II

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算法描述：

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?

解题思路：这道题需要比原题多考虑一个特殊情况，if(nums[left]==nums[mid] && nums[mid] == nums[right]){ left++;right--;}

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

LeetCode-81-Search in Rotated Sorted Array II

标签：leetcode   esc   int   als   com   may   and   特殊情况   12px   

原文地址：https://www.cnblogs.com/nobodywang/p/10345413.html