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。
- 比较A[mid]与target是否相等
- A[mid]==target:返回
midtrue - A[mid]>target or mid
- A[mid]==target:返回
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
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