33. Search in Rotated Sorted Array

最新推荐文章于 2020-08-05 08:57:53 发布

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最新推荐文章于 2020-08-05 08:57:53 发布

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本文介绍了一种在已知数组经过旋转后的环境中查找特定目标值的高效算法。该算法利用了改进的二分查找方法，能够根据数组是否发生旋转来调整搜索范围，确保即使在数组被重新排列的情况下也能找到目标值。文章提供了详细的实现代码，展示了如何根据不同情况选择正确的中间元素进行比较，最终定位到目标值的位置。

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Suppose an array sorted in ascending order 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(vector<int>& nums, int target) {
        int l = 0, r = nums.size()-1;
        while (l <= r) {
            int mid = (l+r) / 2;
            if (target == nums[mid])
                return mid;
            // there exists rotation; the middle element is in the left part of the array
            if (nums[mid] > nums[r]) {
                if (target < nums[mid] && target >= nums[l])
                    r = mid - 1;
                else
                    l = mid + 1;
            }
            // there exists rotation; the middle element is in the right part of the array
            else if (nums[mid] < nums[l]) {
                if (target > nums[mid] && target <= nums[r])
                    l = mid + 1;
                else
                    r = mid - 1;
            }
            // there is no rotation; just like normal binary search
            else {
                if (target < nums[mid])
                    r = mid - 1;
                else
                    l = mid + 1;
            }
        }
        return -1;
    }
};