153. Find Minimum in Rotated Sorted Array QuestionEditorial Solution
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(i.e., 0 1 2 4 5 6 7
might become 4 5 6 7 0 1 2
).
153. Find Minimum in Rotated Sorted Array
我们可能首先想到大擂台赛,故时间复杂度0(n),但题目明显不是考这个,所以我们想到0(log)
首先这个数组旋转,肯定是先一个升序,后一个降序排序,但是根据移动元素的多少分为下列两种:
0 1 2 4 5 6 7
1. 4 5 6 7 0 1 2 移动元素大于一半
2 6 7 0 1 2 4 5 移动元素小于一半
class Solution {
public:
int findMin(vector<int>& nums) {
int l = 0;
int r = nums.size();
int min_ = nums[0];
while(l<r)
{
int mid = (l+r)/2;
if(nums[l] < nums[mid] )
l = mid;
else
r = mid;
if(nums[mid] < min_)
min_ = nums[mid];
}
return min_;
}
};
0 0
- 153. Find Minimum in Rotated Sorted Array QuestionEditorial Solution
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- 153. Find Minimum in Rotated Sorted Array
- 153. Find Minimum in Rotated Sorted Array
- 153. Find Minimum in Rotated Sorted Array
- 153. Find Minimum in Rotated Sorted Array
- 153. Find Minimum in Rotated Sorted Array
- 153. Find Minimum in Rotated Sorted Array
- 153. Find Minimum in Rotated Sorted Array
- 153. Find Minimum in Rotated Sorted Array
- 153. Find Minimum in Rotated Sorted Array
- 153. Find Minimum in Rotated Sorted Array
- 153. Find Minimum in Rotated Sorted Array
- 153. Find Minimum in Rotated Sorted Array
- 153. Find Minimum in Rotated Sorted Array
- 153. Find Minimum in Rotated Sorted Array **
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