Leetcode | Search in Rotated Sorted Array I & II

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 };