[LeetCode] Search in Rotated Sorted Array 解题报告

最新推荐文章于 2020-04-01 23:00:39 发布

原创最新推荐文章于 2020-04-01 23:00:39 发布 · 195 阅读

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本文介绍如何在已知部分排序的数组中查找特定目标值，并提供了优化的二分查找算法来解决此问题。

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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).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

» Solve this problem

[解题思想]
同样是二分，难度主要在于左右边界的确定。需要结合两个不等式：
1. A[m] ? A[left]
2. A[m] ? target
具体逻辑看code。

[Code]

1:       int search(int A[], int n, int target) {   
2:            // Start typing your C/C++ solution below   
3:            // DO NOT write int main() function   
4:            int l = 0, r = n-1;   
5:            while(l<=r)   
6:            {   
7:                 int m = (l+r)/2;   
8:                 if(A[m] == target) return m;   
9:                 if(A[m]>= A[l])   
10:                 {   
11:                      if(A[l]<=target && target<= A[m])   
12:                      r=m-1;   
13:                      else   
14:                      l = m+1;       
15:                 }   
16:                 else   
17:                 {   
18:                      if(A[m] >= target || target>= A[l])   
19:                      r = m-1;    
20:                      else   
21:                      l = m+1;   
22:                 }   
23:            }   
24:            return -1;   
25:       }

Update 08/23/2014
See the comments from reader. Add a graph and also change the code a bit for readability(See highlight code in red).

The general idea is, to use some in-equations to distinguish below 3 conditions, and decide the new range of binary search.

1:       int search(int A[], int n, int target) {    
2:            int l = 0, r = n-1;    
3:            while(l<=r)    
4:            {    
5:                 int m = (l+r)/2;    
6:                 if(A[m] == target) return m;    
7:                 if(A[m]>= A[l])    
8:                 {    
9:                      if(A[l]<=target && target< A[m])    
10:                           r=m-1;    
11:                      else    
12:                           l = m+1;      
13:                 }    
14:                 else    
15:                 {    
16:                      if(A[m]< target && target<=A[r])    
17:                           l = m+1;    
18:                      else    
19:                           r = m-1;   
20:                 }    
21:            }    
22:            return -1;    
23:       }