题目:
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]
).
Find the minimum element.
The array may contain duplicates.
Example 1:
Input: [1,3,5] Output: 1
Example 2:
Input: [2,2,2,0,1] Output: 0
Note:
- This is a follow up problem to Find Minimum in Rotated Sorted Array.
- Would allow duplicates affect the run-time complexity? How and why?
分析:
这道题是153题的延申,153题链接在这里https://www.cnblogs.com/silentteller/p/11163491.html
多了一个限制,也就是允许重复元素出现,所以一次划分后就不会像153题那样出现一个有序数组和一个有可能有序数组的结果。
比如[3,3,3,3,1,3]划分后[3,3,3]和[3,1,3]两个数组均无序,所以要继续对这两个数组进行划分,而且当左右两端元素相同时,不能单纯的认为其内元素都相同,要继续划分,直到出现左端元素小于右端元素为止。
程序和153题相同,因为在判断数组是否有序的时候时严格判断左端值小于右端值的。
程序:
class Solution { public: int findMin(vector<int>& nums) { if(nums.size()==1) return nums[0]; int l = 0; int r = nums.size()-1; return find(nums, l, r); } int find(vector<int>& nums, int l, int r){ if(nums[l] < nums[r] || l==r) return nums[l]; if(r-l == 1) return min(nums[l], nums[r]); int mid = (l+r)/2; return min(find(nums, l, mid-1), find(nums, mid, r)); } };