【leetcode】Search in Rotated Sorted Array

Search in Rotated Sorted Array

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.

 

 

利用二分法，

如果left<A[mid]，说明从左边left开始到mid是有序的

 

如果left>A[mid]，说明右边从mid开始，到right是有序的

 

如果left=A[mid]，则说明left和mid在同一个位置，如果A[mid]!=target则下一步中left=left+1;

 

 1 class Solution {

 2 public:

 3     int search(int A[], int n, int target) {

 4        

 5         int left=0;

 6         int right=n-1;

 7         int mid;

 8         while(left<=right)

 9         {

10             mid=(left+right)/2;

11            

12             if(A[mid]==target) return mid;

13            

14             if(A[left]<A[mid])//left

15             {

16                 if(A[left]<=target&&target<A[mid])

17                 {

18                     right=mid-1;

19                 }

20                 else

21                 {

22                     left=mid+1;

23                 }

24             }

25             else if(A[left]>A[mid])//right

26             {

27                 if(A[mid]<target&&target<=A[right])

28                 {

29                     left=mid+1;

30                 }

31                 else

32                 {

33                     right=mid-1;

34                 }

35             }

36             else

37             {

38                 //最后一个判断语句可以一起放到第一个语句里，if(A[left]<=A[mid])

39                 left=mid+1;

40             }

41            

42         }

43         return -1;

44     }

45 };