Find Minimum in Rotated Sorted Array II

最新推荐文章于 2022-05-26 16:47:57 发布

转载最新推荐文章于 2022-05-26 16:47:57 发布 · 53 阅读

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原文链接：http://www.cnblogs.com/reynold-lei/p/4044984.html

本文探讨了在包含重复元素的旋转排序数组中查找最小元素的算法，详细介绍了处理策略及其对运行时复杂度的影响。

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.

 1 public class Solution {
 2     public int findMin(int[] num) {
 3         if(num == null || num.length == 0) return 0;
 4         int min = Integer.MAX_VALUE;
 5         int str = 0, end = num.length - 1;
 6         while(str <= end){
 7             if(num[str] == num[end]){
 8                 int tmp = num[str];
 9                 min = min > tmp ? tmp : min;
10                 while(str < end && num[str] == tmp) str ++;
11                 while(str < end && num[end] == tmp) end --;
12             }
13             int mid = (str + end) / 2;
14             min = min > num[mid] ? num[mid] : min;
15             if(num[mid] < num[end]){
16                 end = mid;
17             }else if(num[mid] == num[end]){//from mid to end all the number is same
18                 end = mid - 1;
19             }else{
20                 str = mid + 1;
21             }
22         }
23         return min;
24     }
25 }

转载于:https://www.cnblogs.com/reynold-lei/p/4044984.html