154. Find Minimum in Rotated Sorted Array II
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?
Approach #1:
class Solution { public: int findMin(vector<int>& nums) { int len = nums.size(); int l = 0, r = len - 1; while (l <= r) { while (nums[l] == nums[l+1] && r > l) l++; while (nums[r] == nums[r-1] && r > l) r--; int m = l + (r - l) / 2; if (l == r) return nums[l]; if (nums[r] > nums[l]) return nums[l]; if (nums[m] > nums[m+1]) return nums[m+1]; if (nums[m] < nums[m-1]) return nums[m]; if (nums[m] > nums[l]) l = m + 1; else r = m - 1; } return -1; } };
Runtime: 4 ms, faster than 98.95% of C++ online submissions for Find Minimum in Rotated Sorted Array II.
Analysis:
this problem have the same idea with `154. Find Minimum in Rotated Sorted Array I`, the only different is when we meet the same element, we should let l++ or r-- in the legal range.
永远渴望,大智若愚(stay hungry, stay foolish)