<LeetCode OJ> Find Minimum in Rotated Sorted Array【153】

153. Find Minimum in Rotated Sorted Array

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Total Accepted: 73048 Total
Submissions: 209952 Difficulty: Medium

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.

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Find Minimum in Rotated Sorted Array II

//思路首先：遍历一遍时间复杂度是O(N)
class Solution {
public:
    int findMin(vector<int>& nums) {
        int minum=nums[0];
        for(int i=1;i<nums.size();i++)
        {
            if(nums[i]<minum)
                minum=nums[i];
        }
        return minum;
    }
};

//思路首先：第二种方法还是直接遍历，
//按照题目所说数组应该是先上升，在下降，如果下降了，这个值就是最小值
class Solution {
public:
    int findMin(vector<int>& nums) {
        int minum=nums[0];
        for(int i=1;i<nums.size();i++)
        {
            if(nums[i-1]>nums[i])
               { 
                   minum=nums[i];
                   break;
               }
        }
        return minum;
    }
};

//思路首先：第三种方法，二分法
//
class Solution {
public:
    int findMin(vector<int>& nums) {
        if(nums.empty())
            return 0;
        if(nums.size() == 1)
            return nums[0];

        int low = 0, high = nums.size()-1;
        while(low < high && nums[low] > nums[high])
        {
            int mid = low + (high-low)/2;
            if(nums[mid] < nums[low])  
                high = mid;
            else if(nums[mid] == nums[low]) 
                return nums[high]; 
            else
                low = mid+1;
        }
        return nums[low];
    }
};