[Leetcode] Search In Rotated Sorted Array (C++)

题目：

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.

Tag:

Array; Binary Search

体会：

这道题和Find the Minimum In Rotated Sorted Array肯定是有联系啦，毕竟给的都是Rotated Sorted Array这种结构。思路是：从middle处开始左右两部分，一部分是sorted的了，另一部分rotated或者sorted。我们要去分析sorted的那个部分。假设左边是sorted部分，如果A[left] <= target < A[mid]，那么target一定只可能出现在这个部分，otherwise，出现在另外一部分。如果右边是sorted，同理。

 1 class Solution {
 2 public:
 3     int search (int A[], int n, int target) {
 4             int low = 0;
 5             int high = n - 1;
 6
 7             while (low <= high) {
 8                 int mid = low + (high - low) / 2;
 9
10                 if (A[mid] == target) {
11                     return mid;
12                 }
13
14                 if (A[mid] < A[high]) {
15                     // right part is sorted, left is rotated part
16                     if (A[mid] < target && target <= A[high]) {
17                         // target must be in right part
18                         low = mid + 1;
19                     } else {
20                         // target must be left part
21                         high = mid - 1;
22                     }
23                 } else {
24                     // left part is sorted
25                     if (A[low] <= target && target < A[mid]) {
26                         high = mid - 1;
27                     } else {
28                         low = mid + 1;
29                     }
30                 }
31             }
32             return -1;
33     }
34 };