leetcode 33 Search in Rotated Sorted Array-优快云博客

本文链接：https://blog.youkuaiyun.com/CLZHIT/article/details/109590195

本文介绍了一种在旋转排序数组中查找特定元素的方法。该数组原本按升序排列，但在未知位置进行了旋转。通过实现一种改进的二分查找算法，可以在O(log n)的时间复杂度内找到目标值或返回-1表示未找到。

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You are given an integer array nums sorted in ascending order, and an integer target.

Suppose that nums 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]).

If target is found in the array return its index, otherwise, return -1.

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4
Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1
Example 3:

Input: nums = [1], target = 0
Output: -1

Constraints:

1 <= nums.length <= 5000
-10^4 <= nums[i] <= 10^4
All values of nums are unique.
nums is guranteed to be rotated at some pivot.
-10^4 <= target <= 10^4

链接：https://leetcode-cn.com/problems/search-in-rotated-sorted-array

解题思路：

采用二分查找寻找value为target的元素。

因为，数组是递增顺序，可以以某个pivot进行旋转，旋转后的两个子数组也分别是递增的序列。

所以，采用二分法查找时，如果nums[mid] == target，则说明找到了，就返回下标index。

否则，比较nums[first]和nums[mid]的大小，如果nums[first] <= nums[mid]，说明从first到mid是递增序列，如果target的值，刚好在这个递增序列内，last = mid，否则target如果不在这个递增序列内，就说明first的值要更新了，first = mid + 1。

如果nums[first] > nums[mid]，就说明first到mid并不是一个递增序列，它肯定是旋转后的两个序列的交叉情况，这时候，从nums[mid]到nums[last-1]就是递增序列，判断，如果target在这个序列内，则first = mid+1，否则更新last = mid即可。

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int first = 0;
        int last = nums.size();
        while(first < last)
        {
            const int mid = first + (last - first) / 2;
            if(nums[mid] == target)
            {
                return mid;
            }
            if(nums[first] <= nums[mid])
            {
                if(target >= nums[first] && target < nums[mid])
                {
                    last = mid;
                }
                else
                {
                    first = mid + 1;
                }
            }
            else
            {
                if(target > nums[mid] && target <= nums[last-1])
                {
                    first = mid + 1;
                }
                else
                {
                    last = mid;
                }
            }
        }
        return -1;
    }
};