leetcode 33. Search in Rotated Sorted Array 二分查找

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

首先通过二分查找得到最小元素的位置，然后用该位置来进行修正下一个二分查找

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int n=nums.size();
        int lo=0, hi=n-1;
        while(lo<hi){
            int mi=(lo+hi)/2;
            if(nums[mi]>nums[hi])lo=mi+1;
            else hi=mi;
        }
        int rot=lo; // lo==hi is where the smallest element locates
        lo=0,hi=n-1;
        while(lo<=hi){
            int mi=(lo+hi)/2;
            int rm=(mi+rot)%n;
            if(nums[rm]==target)return rm;
            else if(nums[rm]>target)hi=mi-1;
            else lo=mi+1;
        }
        return -1;
    }
};

还有一个很巧妙的做法是判断当前的中位值和目标值是否在同一侧，参考 https://leetcode.com/problems/search-in-rotated-sorted-array/discuss/14435/Clever-idea-making-it-simple：

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int n=nums.size();
        int lo=0, hi=n-1;
        while(lo<=hi){
            int mi=(lo+hi)/2;
            int num=(target<nums[0])==(nums[mi]<nums[0])?nums[mi] // on the same side
                                    :(target>nums[mi]?INT_MAX:INT_MIN); // different side
            if(num==target)return mi;
            if(num>target)hi=mi-1;
            else lo=mi+1;
        }
        return -1;
    }
};