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; }
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