【LeetCode】154. Find Minimum in Rotated Sorted Array II (3 solutions)

Find Minimum in Rotated Sorted Array II

Follow up for "Find Minimum in Rotated Sorted Array": What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

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.

The array may contain duplicates.

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Find Minimum in Rotated Sorted Array对照看，

一共有两处修改。

1、在无重复元素时，首尾元素相等代表指向同一个位置，因此程序直接返回即可。

然而当存在重复元素时，该条件并不能表示指向同一个位置，因此

nums[low] > nums[high]

改为

nums[low] >= nums[high]

2、在无重复元素时，中间元素与首元素相等，表示一共只有两个元素，low与high各指向一个。

由于while循环中限制的大小关系，因此返回nums[high]即为最小值。

然而当存在重复元素时，该条件并不能表示一共只有low和high指向的两个元素，

而是说明low指向的元素重复了，因此删除其一，low ++即可。

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 parthigh = mid;else if(nums[mid] == nums[low])low ++;elselow = mid+1;}return nums[low];}
};

转载于:https://www.cnblogs.com/ganganloveu/p/4081483.html

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