[LeetCode] 154. Find Minimum in Rotated Sorted Array II 寻找旋转有序数组的最小值之二
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?
这道寻找旋转有序重复数组的最小值是之前那道 Find Minimum in Rotated Sorted Array 的拓展,当数组中存在大量的重复数字时,就会破坏二分查找法的机制,将无法取得 O(lgn) 的时间复杂度,又将会回到简单粗暴的 O(n),比如这两种情况:{2, 2, 2, 2, 2, 2, 2, 2, 0, 1, 1, 2} 和 {2, 2, 2, 0, 2, 2, 2, 2, 2, 2, 2, 2},可以发现,当第一个数字和最后一个数字,还有中间那个数字全部相等的时候,二分查找法就崩溃了,因为它无法判断到底该去左半边还是右半边。这种情况下,将右指针左移一位(或者将左指针右移一位),略过一个相同数字,这对结果不会产生影响,因为只是去掉了一个相同的,然后对剩余的部分继续用二分查找法,在最坏的情况下,比如数组所有元素都相同,时间复杂度会升到 O(n),参见代码如下:
解法一:
class Solution {
public:
int findMin(vector<int>& nums) {
int left = , right = (int)nums.size() - ;
while (left < right) {
int mid = left + (right - left) / ;
if (nums[mid] > nums[right]) left = mid + ;
else if (nums[mid] < nums[right]) right = mid;
else --right;
}
return nums[right];
}
};
跟之前那道 Find Minimum in Rotated Sorted Array 一样,还是可以用分治法 Divide and Conquer 来解,还是由热心网友 howard144 提供,不过写法跟之前那道略有不同,只有在 nums[start] < nums[end] 的时候,才能返回 nums[start],等于的时候不能返回,比如 [3, 1, 3] 这个数组,或者当 start 等于 end 成立的时候,也可以直接返回 nums[start],后面的操作跟之前那道题相同,每次将区间 [start, end] 从中间 mid 位置分为两段,分别调用递归函数,并比较返回值,每次取返回值较小的那个即可,参见代码如下:
解法二:
class Solution {
public:
int findMin(vector<int>& nums) {
return helper(nums, , (int)nums.size() - );
}
int helper(vector<int>& nums, int start, int end) {
if (start == end) return nums[start];
if (nums[start] < nums[end]) return nums[start];
int mid = (start + end) / ;
return min(helper(nums, start, mid), helper(nums, mid + , end));
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/154
类似题目:
Find Minimum in Rotated Sorted Array
参考资料:
https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/
LeetCode All in One 题目讲解汇总(持续更新中...)
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