leetcode_153_Find Minimum in Rotated Sorted Array

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Find Minimum 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).

Find the minimum element.

You may assume no duplicate exists in the array.

测试例子：[1,2]      [2,1]    [4,5,6,7,8,9,10,11,1,2,3]

//vs2012测试代码
//方法一：二分查找
//主要思想是二分查找，对于num[left .. right]数组，命mid=(left+right)/2,那么存在一下三种情况：
//1)num[left]<num[right]： 这时，整个数组是有序的，因此直接返回num[left]即可；
//2)num[left]>=num[right] && num[left]>num[mid]：这时，num[mid..right]有序，那么最小值一定出现在num[left..mid]之中，抛弃num[mid+1..right],则取right=mid;
//3)num[left]>=num[right] && num[left]<=num[mid] : 这时,num[left..mid]是有序的，所以最小值一点个出现在num[mid+1..right]中,抛弃num[left..mid],则取left=mid+1;

#include<iostream>
#include<vector>
#include<algorithm>

using namespace std;

#define N 4

class Solution {
public:
    int findMin(vector<int> &num) 
	{
		int size = num.size();
		int left = 0 , right = size-1;
		while( left<right)
		{
			//all sorted
			if( num[left] < num[right])
				break;
			else
			{
				int mid = left + (right-left)/2;
				//Sorted from mid to right  
				//The smallest must in num[left..mid] 
				if( num[mid] < num[left])
					right = mid; 

				//Sorted from beg to mid  
                //The smallest must in num[mid+1..end] 
				else
					left = mid+1;
			}
		}
		return num[left];
    }
};

int main()
{
	vector<int> num;
	int a;
	for(int i=0; i<N; i++)
	{
		cin>>a;
		num.push_back(a);
	}
	Solution lin;
	cout<<lin.findMin(num)<<endl;

	return 0;
}

//方法一：自测Accepted
//主要思想是二分查找，对于num[left .. right]数组，命mid=(left+right)/2,那么存在一下三种情况：
//1)num[left]<num[right]： 这时，整个数组是有序的，因此直接返回num[left]即可；
//2)num[left]>=num[right] && num[left]>num[mid]：这时，num[mid..right]有序，那么最小值一定出现在num[left..mid]之中，抛弃num[mid+1..right],则取right=mid;
//3)num[left]>=num[right] && num[left]<=num[mid] : 这时,num[left..mid]是有序的，所以最小值一点个出现在num[mid+1..right]中,抛弃num[left..mid],则取left=mid+1;

class Solution {
public:
    int findMin(vector<int> &num) 
	{
		int size = num.size();
		int left = 0 , right = size-1;
		while( left<right)
		{
			//all sorted
			if( num[left] < num[right])
				break;
			else
			{
				int mid = left + (right-left)/2;
				//Sorted from mid to right  
				//The smallest must in num[left..mid] 
				if( num[mid] < num[left])
					right = mid; 

				//Sorted from beg to mid  
                //The smallest must in num[mid+1..end] 
				else
					left = mid+1;
			}
		}
		return num[left];
    }
};

//方法二:其他版本
class Solution {
public:
    int findMin(vector<int> &num) 
	{
		int length = num.size()-1;
		int left=0 , right=length;
		while( left<=right)
		{
			int mid = left + (right-left)/2;
			if( num[mid] > num[length])
			{
				left = mid + 1;
			}
			else
			{
				right = mid - 1;
			}
		}
		return num[left];
    }
};

//方法三：书中版本
class Solution {
public:
    int findMin(vector<int> &num) 
	{
		int left = 0;
		int right = num.size()-1;
		int mid = left;
		while(num[left]>num[right])
		{
		    if(right-left == 1)
		    {
		        mid = right;
		        break;
		    }
		    
		    mid = (right+left)/2;
		    if(num[left] > num[mid])
		        right = mid;
		    else
		        left = mid;
		}
		return num[mid];
    }
};