[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 = 0;
            int high = n - 1;
            
            while (low <= high) {
                int mid = low + (high - low) / 2;
                
                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 + 1;
                    } else {
                        // target must be left part
                        high = mid - 1;
                    }
                } else {
                    // left part is sorted
                    if (A[low] <= target && target < A[mid]) {
                        high = mid - 1;
                    } else {
                        low = mid + 1;
                    }
                }
            }
            return -1;
    }
};