[LeetCode] (medium) 33. Search in Rotated Sorted Array

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

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

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有序数列查找外加logn复杂度，那么几乎可以确定是二分查找，但是因为是个旋转数列，所以是二分查找的变形。

在针对[mid]和target不同大小情况下如何移动两端界限是问题的关键，看似复杂，实际上只要把握好二分查找的关键思想——“两端界限之间的区域是可能的解空间（目标不可能落在两端界限之外）”——只要每一次的移动都确保了这个二分不变性即可。以这个思想为基础，可以针对[mid]大于target的情况可以将mid指针与tar虚指针（target存在的合法位置）的位置关系分为两类：1、两者在同一个有序序列中。2、两者跨立在两个各自有序序列中。判断方式很直观，移动方式也很容易想。小于的情况同理。

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int lo = 0;
        int hi = nums.size()-1;
        int mid;
        while(lo <= hi){
            mid = (lo+hi)/2;
            //cout << "lo: " << lo << " mid: " << mid << " hi: " << hi << " [mid]: " << nums[mid] << endl;
            if(nums[mid] == target) return mid;
            else if(nums[mid] > target){    //mid和tar的位置共有三种可能
                if(nums[lo] > target  && nums[mid] >= nums[lo]){    //mid在左侧，tar在右侧
                    lo = mid+1;
                }else{  //mid与tar在同侧
                    hi = mid-1;
                }
            }else{
                if(nums[hi] < target && nums[mid] <= nums[hi]){
                    hi = mid-1;
                }else{
                    lo = mid+1;
                }
            }
        }
        return -1;
    }
};