leetcode33. Search in rotated sorted array

标签：lex   思路   array   asc   The   简单   bec   ret   ted   

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;
    }
};

leetcode33. Search in rotated sorted array

标签：lex   思路   array   asc   The   简单   bec   ret   ted   

原文地址：https://www.cnblogs.com/yaoyudadudu/p/9120177.html