版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/mike_learns_to_rock/article/details/47188249
Suppose a sorted array 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).
Find the minimum element.
You may assume no duplicate exists in the array.
参考leetcode solution;
对于数组A,AL代表最左边的元素,AR代表最右边的元素;
如果没旋转,AL
int findMin(vector<int>& nums) {
int n=nums.size();
if(n==1) return nums[0];
int first=0,last=n-1;
//consider special conditions
//if(nums[first]<nums[last]) return nums[first];
while(first<last){
int mid=(first+last)/2;
if(nums[last]<nums[mid]) first=mid+1;
else last=mid;
}
return nums[last];
}
如果有重复元素呢?
最坏情况时间复杂度O(n);
直接去除数组初始的重复元素(nums[first]==nums[last]),后面和没重复的一样了!见下一篇博客分析!