友情支持
如果您觉得这个笔记对您有所帮助，看在D瓜哥码这么多字的辛苦上，请友情支持一下，D瓜哥感激不尽，😜
|
|
有些打赏的朋友希望可以加个好友，欢迎关注D 瓜哥的微信公众号，这样就可以通过公众号的回复直接给我发信息。
公众号的微信号是: jikerizhi。因为众所周知的原因，有时图片加载不出来。 如果图片加载不出来可以直接通过搜索微信号来查找我的公众号。 |
153. Find Minimum in Rotated Sorted Array
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.
You may assume no duplicate exists in the array.
Example 1:
Input: [3,4,5,1,2] Output: 1
Example 2:
Input: [4,5,6,7,0,1,2] Output: 0
思路分析
最省事的方式当然是遍历，但是，中等难度的题目不可能一个遍历就解决了。
针对有序数组，首先想到的就是二分查找。但是，这里的二分查找有需要做些特别处理：要判断最低点在哪个区间。
-
一刷
-
二刷
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
/**
* Runtime: 0 ms, faster than 100.00% of Java online submissions for Find Minimum in Rotated Sorted Array.
* Memory Usage: 39.8 MB, less than 12.50% of Java online submissions for Find Minimum in Rotated Sorted Array.
*
* @author D瓜哥 · https://www.diguage.com
* @since 2020-04-25 22:37
*/
public int findMin(int[] nums) {
if (nums.length == 1) {
return nums[0];
}
int left = 0;
int right = nums.length - 1;
if (nums[left] < nums[right]) {
return nums[left];
}
int mid = 0;
while (left <= right) {
mid = left + (right - left) / 2;
if (mid + 1 < nums.length && nums[mid] > nums[mid + 1]) {
return nums[mid + 1];
}
if (mid - 1 >= 0 && nums[mid - 1] > nums[mid]) {
return nums[mid];
}
if (nums[left] < nums[mid]) {
left = mid + 1;
} else {
right = mid - 1;
}
}
return -1;
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
/**
* @author D瓜哥 · https://www.diguage.com
* @since 2024-09-19 20:51:15
*/
public int findMin(int[] nums) {
int left = 0, right = nums.length - 1;
while (left < right) {
int mid = left + (right - left) / 2;
if (nums[mid]< nums[right]) {
right = mid;
} else {
left = mid +1;
}
}
return nums[left];
}