二进制搜索可以在旋转排序列表中查找旋转点 [英] Binary search to find the rotation point in a rotated sorted list

问题描述

我有一个排序的列表，它被旋转，并希望在该列表上进行二进制搜索以找到最小元素。

I have a sorted list which is rotated and would like to do a binary search on that list to find the minimum element.

假设初始列表是{1 ，2,3,4,5,6,7,8}
旋转列表可以像{5,6,7,8,1,2,3,4}

Lets suppose initial list is {1,2,3,4,5,6,7,8} rotated list can be like {5,6,7,8,1,2,3,4}

正常二进制搜索在这种情况下不起作用。任何想法如何做到这一点。

Normal binary search doesn't work in this case. Any idea how to do this.

- 修改

我有另一个条件。如果列表没有排序怎么办？

I have one another condition. What if the list is not sorted??

推荐答案

二进制搜索算法的一个修改就是你需要的;以下是完整可运行Java的解决方案（请参阅 Serg的答案用于Delphi实现，而 tkr的答案，用于视觉解释算法）。

A slight modification on the binary search algorithm is all you need; here's the solution in complete runnable Java (see Serg's answer for Delphi implementation, and tkr's answer for visual explanation of the algorithm).

import java.util.*;
public class BinarySearch {
    static int findMinimum(Integer[] arr) {
        int low = 0;
        int high = arr.length - 1;
        while (arr[low] > arr[high]) {
            int mid = (low + high) >>> 1;
            if (arr[mid] > arr[high]) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }
        return low;
    }
    public static void main(String[] args) {
        Integer[] arr = { 1, 2, 3, 4, 5, 6, 7 };
        // must be in sorted order, allowing rotation, and contain no duplicates

        for (int i = 0; i < arr.length; i++) {
            System.out.print(Arrays.toString(arr));
            int minIndex = findMinimum(arr);
            System.out.println(" Min is " + arr[minIndex] + " at " + minIndex);
            Collections.rotate(Arrays.asList(arr), 1);
        }
    }
}

打印：

[1, 2, 3, 4, 5, 6, 7] Min is 1 at 0
[7, 1, 2, 3, 4, 5, 6] Min is 1 at 1
[6, 7, 1, 2, 3, 4, 5] Min is 1 at 2
[5, 6, 7, 1, 2, 3, 4] Min is 1 at 3
[4, 5, 6, 7, 1, 2, 3] Min is 1 at 4
[3, 4, 5, 6, 7, 1, 2] Min is 1 at 5
[2, 3, 4, 5, 6, 7, 1] Min is 1 at 6

另请参阅

• Java Collections.rotate（）与数组不起作用$
  
  
  
  
  • 解释为什么 Integer [] 而不是 int []
  
  

  See also

  • Java Collections.rotate() with an array doesn’t work
    • Explains why Integer[] instead of int[]
    
    
    • 解释为什么 >>> 1 而不是 / 2
    
    
    • Explains why >>> 1 instead of / 2

    请注意，重复项不可能在 O（log N）。考虑以下位数组，它由许多 1 和一个 0 组成：

    Note that duplicates makes it impossible to do this in O(log N). Consider the following bit array consisting of many 1, and one 0:

      (sorted)
      01111111111111111111111111111111111111111111111111111111111111111
      ^
    
      (rotated)
      11111111111111111111111111111111111111111111101111111111111111111
                                                   ^
    
      (rotated)
      11111111111111101111111111111111111111111111111111111111111111111
                     ^
    

    可以在 N 中旋转，并将 0 在 O（log N）是不可能的，因为没有办法告诉它是否在中间的左侧或右侧。

    This array can be rotated in N ways, and locating the 0 in O(log N) is impossible, since there's no way to tell if it's in the left or right side of the "middle".

    
    

    我有另一个条件。如果列表没有排序，该怎么办？

    I have one another condition. What if the list is not sorted??

    然后，除非你想先排序，然后从那里进行排序进行线性搜索以找到最小值。

    Then, unless you want to sort it first and proceed from there, you'll have to do a linear search to find the minimum.

    
    
    • <一个href =http://en.wikipedia.org/wiki/Selection_algorithm#Linear_minimum.2Fmaximum_algorithms =noreferrer>维基百科|选择算法|线性最小/最大算法
    
    
    • Wikipedia | Selection algorithm | Linear minimum/maximum algorithms

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