Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:
- [4,5,6,7,0,1,2] if it was rotated 4 times.
- [0,1,2,4,5,6,7] if it was rotated 7 times.
Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].
Given the sorted rotated array nums of unique elements, return the minimum element of this array.
You must write an algorithm that runs in O(log n) time.
class Solution {
public:
int findMin(vector<int>& nums) {
//O(logn 的时间复杂度 感觉是二分法的检索方式) left>right 的话 最小值就在这里面 这样二分检索
//二分检索 中间的边界比较麻烦
int n=nums.size();
if(nums[0]<=nums[n-1]) return nums[0];
int left=0,right=n-1;
while(left<right-1)
{
int mid=(left+right)/2;
if(nums[left]>nums[mid])
{
//left=left;
right=mid;
}
else if(nums[mid]>nums[right])
{
left=mid;
//right=right
}
else if(nums[left]==nums[mid])
{
left=mid+1;
}
else if(nums[right]==nums[mid])
{
right=mid-1;
}
}
return min(nums[right],nums[left]);
}
};