[LeetCode] Search in Rotated Sorted Array

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本文介绍了一种在旋转排序数组中查找目标值的高效算法。该算法利用二分搜索原理，在O(log n)时间内找到目标值的索引，若不存在则返回-1。文章通过实例展示了数组旋转的特点，并提供了一个C++实现的例子。

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.

class Solution {
public:
    int search(int A[], int n, int target) {
        int left = 0, right = n-1;
        while(left <= right)
        {
            int mid = left + (right - left) / 2;
            if(A[mid] == target)    return mid;
            if(A[mid] >= A[left])
            {
                if(A[left] <= target && target < A[mid])
                    right = mid - 1;
                else
                    left = mid + 1;
            }
            else
            {
                if(A[mid] < target && target <= A[right])
                    left = mid + 1;
                else
                    right = mid - 1;
            }
        }
        return -1;
    }
};