033. Search in Rotated Sorted Array
难度Medium
刷题内容
原题连接
- https://leetcode.com/problems/search-in-rotated-sorted-array/
内容描述
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).
You are given a target value to search. If found in the array return its index, otherwise return -1.
You may assume no duplicate exists in the array.
Your algorithm's runtime complexity must be in the order of O(log n).
Example 1:
Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4
Example 2:
Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1
思路1 **- 时间复杂度: O(n)*- 空间复杂度: O(1)***
第一个方法是直接遍历数组，找到返回数组下标，找不到就返回-1
class Solution {
public:
int search(vector<int>& nums, int target) {
for(int i = 0;i < nums.size();++i)
if(nums[i] == target)
return i;
return -1;
}
};
思路2 **- 时间复杂度: O(lgn)- 空间复杂度: O(1)***
第二个方法是用二分法找到旋转轴，再用二分法找到目标数
class Solution {
public:
int search(vector<int>& nums, int target) {
int i = 0,j = nums.size() - 1;
if(!nums.size())
return -1;
while(i < j - 1)
{
int mid = (i + j) / 2;
if(nums[i] < nums[mid])
i = mid;
else
j = mid;
//cout << i << j << endl;
}
if(nums[i] <= nums[j])
j = i;
//cout << j;
auto pos = lower_bound(nums.begin(),nums.begin() + j,target);
if(pos != nums.end() && (*pos) == target)
return pos - nums.begin();
pos = lower_bound(nums.begin() + j,nums.end(),target);
if(pos != nums.end() && (*pos) == target)
return pos - nums.begin();
return -1;
}
};
