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Leetcode | Search in Rotated Sorted Array I & II

2024-07-05 00:06:52   226人阅读

Search in Rotated Sorted Array I 

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.

Have you been asked this question in an interview?

二分查找。

1. 如果A[mid]>A[l]，也就是A[l...mid]是递增序。如果target>=A[l]&&target<A[mid]，那么target只可能在[l...mid-1]这个区间上，否则在[mid+1...h]这个区间。

2. 如果A[mid]<=A[l],那么A[mid...h]是递增序。如果target>A[mid] && target<=A[h]，那么target只可能在[mid+1...h]这个区间，否则在[l...mid-1]这个区间。

 1 class Solution {
 2 public:
 3     int search(int A[], int n, int target) {
 4         if (n == 0) return -1;
 5         int l = 0, h = n - 1, mid;
 6         
 7         while (l <= h) {
 8             mid = (l + h) / 2;
 9             if (A[mid] == target) return mid;
10             if (A[mid] >= A[l]) {
11                 if (target >= A[l] && target < A[mid]) {
12                     h = mid - 1;
13                 } else {
14                     l = mid + 1;
15                 }
16             } else {
17                 if (target > A[mid] && target <= A[h]) {
18                     l = mid + 1;
19                 } else {
20                     h = mid - 1;
21                 }
22             }
23         }
24         
25         return -1;
26     }
27 };

Search in Rotated Sorted Array II

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.

有了重复数。那么前面的情况，只对应于A[mid]>A[l]和A[mid]<A[l]是成立的。对于A[mid]==A[l]，我们无法确定该数是在左区间还是右区间，反正一定不会是在A[l]上，所以l++。

最坏情况就是整个数组都是同一个数，target不在数组中，这样每次都只能l++，算法复杂度是O(n)。

 1 class Solution {
 2 public:
 3     bool search(int A[], int n, int target) {
 4        if (n == 0) return false;
 5         int l = 0, h = n - 1, mid;
 6         
 7         while (l <= h) {
 8             mid = (l + h) / 2;
 9             if (A[mid] == target) return true;
10             if (A[mid] == A[l]) {
11                 l++;
12             } else if (A[mid] > A[l]) {
13                 if (target >= A[l] && target < A[mid]) {
14                     h = mid - 1;
15                 } else {
16                     l = mid + 1;
17                 }
18             } else {
19                 if (target > A[mid] && target <= A[h]) {
20                     l = mid + 1;
21                 } else {
22                     h = mid - 1;
23                 }
24             }
25         }
26         
27         return false;
28     }    
29 };

 


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