Find Minimum in Rotated Sorted Array
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 rotated4
times.[0,1,2,4,5,6,7]
if it was rotated7
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.
Example 1:
Input: nums = [3,4,5,1,2] Output: 1 Explanation: The original array was [1,2,3,4,5] rotated 3 times.
Example 2:
Input: nums = [4,5,6,7,0,1,2] Output: 0 Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.
Example 3:
Input: nums = [11,13,15,17] Output: 11 Explanation: The original array was [11,13,15,17] and it was rotated 4 times.
Constraints:
n == nums.length
1 <= n <= 5000
-5000 <= nums[i] <= 5000
- All the integers of
nums
are unique. nums
is sorted and rotated between1
andn
times.
题解
从题目可以知道给定的数组是经过一个递增的数组经过多轮rotate转化而来的,每轮rotate将[a[0], a[1], a[2], ..., a[n-1]]转化成[a[n-1], a[0], a[1], a[2], ..., a[n-2]]。
因此,经过多轮rotate后的数组nums一定有以下关系:nums[0]~nums[i]递增;nums[i+1]~nums[n-1]递增。因此想要找到数组nums的最小值,必须要找到下标i所在位置。题目要求时间复杂度为O(logn),我们可以利用二分法进行搜索。
搜索方案如下:
while(l<=r){
mid=l+(r-l)/2;
if(nums[mid]<nums[mid-1]&&nums[mid]<nums[mid+1]){
return nums[mid];
}else{
if(nums[mid]>nums[n-1]){
l=mid+1;
}else{
r=mid-1;
}
}
}
完整代码
class Solution {
public:
int findMin(vector<int>& nums) {
// nums[i:j] ascending ; nums[j+1:n] ascending
int n=nums.size();
if(n==1){
return nums[0];
}else if(n<3){
return min(nums[0],nums[1]);
}
int l=1,r=n-2,mid,ans;
while(l<=r){
mid=l+(r-l)/2;
if(nums[mid]<nums[mid-1]&&nums[mid]<nums[mid+1]){
return nums[mid];
}else{
if(nums[mid]>nums[n-1]){
l=mid+1;
}else{
r=mid-1;
}
}
}
return min(nums[0],nums[n-1]);
}
};