【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 };