【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 = 0;
        int end = nums.size() - 1;
        while (start <= end) {
            int mid = (start + end) / 2;
            if (nums[mid] == target) 
                return mid;
            if (nums[start] <= nums[mid]) {
                if (nums[start] <= target && target <= nums[mid])
                    end = mid - 1;
                else 
                    start = mid + 1;
            } else {
                if (nums[mid] <= target && target <= nums[end])
                    start = mid + 1;
                else
                    end = mid - 1;
            }
        }
        
        return -1;   
    }
};

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 = 0;
        int right = nums.size() - 1;
        while (left < right) {
            int mid = (left + right) / 2;
            if (nums[mid] > target) {
                if (nums[left] <= target || nums[left] > nums[mid])
                    right = mid;
                else
                    left = mid + 1;
            } else if (nums[mid] == target) {
                left = mid;
                break;
            } else {
                if (nums[right] >= target || nums[mid] > nums[right])
                    left = mid + 1;
                else
                    right = mid;
            }
        }
        
        if (nums[left] == target)
            return left;
        else 
            return -1;
    }
};