【LeetCode】153. Find Minimum in Rotated Sorted Array (3 solutions)

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.

 

Search in Rotated Sorted Array,Search in Rotated Sorted Array II,Find Minimum in Rotated Sorted Array II对照看

解法一:暴力解法,直接使用algorithm库中的求最小元素函数

需要遍历整个vector

class Solution {
public:
    int findMin(vector<int> &num) {
        if(num.empty())
            return 0;
        vector<int>::iterator iter = min_element(num.begin(), num.end());
        return *iter;
    }
};

 

解法二:利用sorted这个信息。如果平移过,则会出现一个gap,也就是从最大元素到最小元素的跳转。如果没有跳转,则说明没有平移。

比上个解法可以省掉不少时间,平均情况下不用遍历vector了。

class Solution {
public:
    int findMin(vector<int> &num) {
        if(num.empty())
            return 0;
        else if(num.size() == 1)
            return num[0];
        else
        {
            for(vector<int>::size_type st = 1; st < num.size(); st ++)
            {
                if(num[st-1] > num[st])
                    return num[st];
            }
            return num[0];
        }
    }
};

 

解法三:二分查找

Search in Rotated Sorted Array对照来看

Search in Rotated Sorted Array题中是二分查找最大值,而本题是二分查找最小值。

class Solution {
public:
    int findMin(vector<int>& nums) {
        if(nums.empty())
            return 0;
        if(nums.size() == 1)
            return nums[0];
        int n = nums.size();
        int low = 0;
        int high = n-1;
        while(low < high && nums[low] > nums[high])
        {
            int mid = low + (high-low)/2;
            if(nums[mid] < nums[low])   // mid is in second part
                high = mid;
            else if(nums[mid] == nums[low]) // since (low<high)-->(low+1==high)
                return nums[high];  // nums[low]>nums[high]
            else
                low = mid+1;
        }
        return nums[low];
    }
};

posted @ 2014-11-07 15:25  陆草纯  阅读(685)  评论(1编辑  收藏  举报