【LeetCode】033. Search in Rotated Sorted Array

题目：

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.

题解：

Solution 1 

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int n = nums.size();
        if(n < 1)
            return -1;
        int begin = 0, end = n;
        while(begin < end){
            int mid = begin + (end - begin) / 2;
            if(nums[mid] == target)
                return mid;
            if(nums[begin] <= nums[mid]){
                if(nums[begin] <= target && target < nums[mid])
                    end = mid;
                else
                    begin = mid + 1;
            } else {
                if(nums[mid] < target && target <= nums[end - 1])
                    begin = mid + 1;
                else
                    end = mid;
            }
        }
        return -1;
    }
};

 

Solution 2 

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int n = nums.size();
        if(n < 1)
            return -1;
        int begin = 0, end = n - 1;
        while(begin < end){
            int mid = begin + (end - begin) / 2;
            if(nums[mid] > nums[end])
                begin = mid + 1;
            else
                end = mid;
        }
        int pivot = begin;
        begin = 0, end = n - 1;
        while(begin <= end){
            int mid = begin + (end - begin) / 2;
            int realmid = (mid + pivot) % n;
            if(nums[realmid] == target)
                return realmid;
            else if(nums[realmid] < target)
                begin = mid + 1;
            else
                end = mid - 1;
        }
        return -1;
    }
};