作者:流纸香p_455 | 来源:互联网 | 2023-10-12 08:32
Followupfor"SearchinRotatedSortedArray":Whatif duplicates areall
Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?
Would this affect the run-time complexity? How and why?
Write a function to determine if a given target is in the array.
题意:查询给定目标值是否在数组中,是search in rotated sorted array的扩展。
思路:因为是上一题的扩展,所以很正常的想到,上题的思路,在本题是否合适。结果发现不能强行的套上去。如:1,1,3,1,target=3,若是按照原来的解法,则应该是左边有序,然而不行;若是if条件判断中间等号,为右边有序,也显然不对。参考了Grandyang的博客才知道,只需在判断上加上判断中间值是否等于最右端值,若是,hi向左移动一个,直到不相等。这样就可以继续保持上题的解法了,具体代码如下:
1 class Solution {
2 public:
3 bool search(int A[], int n, int target)
4 {
5 if(n==0) return false;
6 int lo=0,hi=n-1;
7 while(lo<=hi)
8 {
9 int mid=(lo+hi)/2;
10 if(A[mid]==target)
11 return true;
12 else if(A[mid]<A[hi])
13 {
14 if(A[mid]=target)
15 lo=mid+1;
16 else
17 hi=mid-1;
18 }
19 else if(A[mid]>A[hi])
20 {
21 if(A[lo]<=target&&A[mid]>target)
22 hi=mid-1;
23 else
24 lo=mid+1;
25 }
26 else //重点理解
27 hi--;
28 }
29 return false;
30 }
31 };