leetcode33. Search in rotated sorted array

问题描述：

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.

Your algorithm's runtime complexity must be in the order of O(log n).

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

 

解题思路：

这道题要求用logN的方法解决，我们就可以开始考虑二分搜索了。

这道题也不能简简单单的就用二分搜索了，因为并不是单调的排序，在某个不知道的地方，它rotate了一下。

这里就需要我们观察！

通过给的例子：[0,1,2,4,5,6,7] -> [4,5,6,7,0,1,2]

可以看出可能有以下rotate：

[1,2,4,5,6,7,0]

[2,4,5,6,7,0,1]

[4,5,6,7,0,1,2]

[5,6,7,0,1,2,4]

[6,7,0,1,2,4,5]

[7,0,1,2,4,5,6]

我们可以看出：当中间的数字小于最右时，右半部分为有序； 当中间的数字大于最左时，左半部分为有序

可以用这个与二分搜索相结合。

代码：

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

 

转载于:https://www.cnblogs.com/yaoyudadudu/p/9120177.html