【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.

与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

与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];
    }
};