[leetcode] 154. Find Minimum in Rotated Sorted Array II 解题报告
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题目链接:https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/
Follow up for "Find Minimum in Rotated Sorted Array":
What if duplicates are allowed?Would this affect the run-time complexity? How and why?
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.
The array may contain duplicates.
思路:和上题一样,多了重复元素。
我们设置两个指针left = 0,right = length - 1;
用二分查找来找到右半有序数组的开始位置。由旋转数组的性质可知,在右半个数组的所有值都小于等于左半数组。每次二分会有3种情况:
1. nums[mid] < nums[right],这种情况是当前的数小于右边数组的一个数,说明当前的位置肯定是在右半数组中,因此可以往左查找[left, mid]
2. nums[mid] > nums[right],这种情况是当前的数大于右边数组的一个数,说明当前的位置在左半数组中,因此要查找右半数组第一个数的位置可以向右搜索[mid+1, right]
3. nums[mid] = nums[right],这种情况下无法判断当前位置是在左边还是右边,但是让右指针左移一位并不会影响到最小值,因为nums[mid] = nums[right], 因此搜索[left, right-1]也就是说多了重复元素比不含重复元素多了一个判断就是当nums[mid] = nums[right]时让right--;这种方法比《剑指offer》上的方法简洁了无数倍,感谢“喜刷刷”博主优雅的解决方案!
代码如下:
class Solution {public: int findMin(vector<int>& nums) { int len = nums.size(), left = 0, right = len-1; while(left <= right) { int mid = (left + right)/2; if(nums[mid] > nums[right]) left = mid+1; else if(nums[mid] < nums[right]) right = mid; else right--; } return nums[left]; }};参考:http://bangbingsyb.blogspot.com/2014/11/leecode-find-minimum-in-rotated-sorted.html
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