leetcode学习篇七——Find Minimum in Rotated Sorted Array
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试了一些pick one功能,随机选取了一道题,题目如下:
153. Find Minimum in Rotated Sorted Array
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.
难度: Medium 通过率:37.9%
这道题最简单的想法就是直接顺序遍历,因为数组原先是排序好的,只经过一次翻转,所以找到某个元素比其前一个元素小即是最小值,时间复杂度:O(n),代码实现如下:
class Solution {public: int findMin(vector<int>& nums) { for(int i = 1; i < nums.size(); i++) { if(nums[i] < nums[i-1]) { return nums[i]; } } return nums[0]; }};
然而这样效率是很低的,所以考虑能否尝试其他做法。利用类似于二分查找的方法寻找最小值,具体实现如下,时间复杂度:O(logn)
class Solution {public: int findMin(vector<int>& nums) { int l = 0, r = nums.size()-1 ; while(l < r) { if(nums[l] < nums[r]) return nums[l]; int mid = (l + r) / 2; if(nums[mid] > nums[r]) { l = mid + 1; } else { r = mid; } } return nums[l]; }};
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