[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。

 class Solution {
 public:
     int search (int A[], int n, int target) {
             int low = ;
             int high = n - ;

             while (low <= high) {
                 int mid = low + (high - low) / ;

                 if (A[mid] == target) {
                     return mid;
                 }

                 if (A[mid] < A[high]) {
                     // right part is sorted, left is rotated part
                     if (A[mid] < target && target <= A[high]) {
                         // target must be in right part
                         low = mid + ;
                     } else {
                         // target must be left part
                         high = mid - ;
                     }
                 } else {
                     // left part is sorted
                     if (A[low] <= target && target < A[mid]) {
                         high = mid - ;
                     } else {
                         low = mid + ;
                     }
                 }
             }
             return -;
     }
 };