【Leetcode】【Hard】Search in Rotated Sorted Array

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.

解题：

二分搜索的应用，可以有两种思路：

（1）将数列从mid处划分，按照前半段序列顺序正常，或者不正常，进行条件判断，达到log(n)的二分效果；

（2）将数列从mid处划分，比较mid处元素和target，按照mid元素大，或者mid元素小，进行条件判断，达到log(n)的二分效果；

代码实现1：

代码简介，灵活的运用的二分的思想，将正常/非正常作为二分标准，而不是mid值大还是小。

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

         return -;
     }
 };

代码实现2：

思路更常规，虽然代码不如1简介，但是逻辑更好理解。

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

         if (nums[left] == target)
             return left;
         else
             return -;
     }
 };