leetcode.321. Create Maximum Number

最新推荐文章于 2023-02-25 22:41:55 发布

happyxuma1991

最新推荐文章于 2023-02-25 22:41:55 发布

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本文介绍了一种算法，该算法接收两个数字数组，并从中选择元素组成一个长度为k的最大可能数值。算法保持每个输入数组内数字的相对顺序不变，并通过优化时间和空间复杂度实现高效处理。

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Given two arrays of length m and n with digits 0-9 representing two numbers. Create the maximum number of length k <= m + n from digits of the two. The relative order of the digits from the same array must be preserved. Return an array of the k digits. You should try to optimize your time and space complexity.

Example 1:

nums1 = [3, 4, 6, 5]
nums2 = [9, 1, 2, 5, 8, 3]
k = 5
return [9, 8, 6, 5, 3]

Example 2:

nums1 = [6, 7]
nums2 = [6, 0, 4]
k = 5
return [6, 7, 6, 0, 4]

Example 3:

nums1 = [3, 9]
nums2 = [8, 9]
k = 3
return [9, 8, 9]

class Solution {
public:
	vector<int> maxNumber(vector<int>& nums1, vector<int>& nums2, int k) {
		int size1 = nums1.size();
		int size2 = nums2.size();
		vector<int> res(k);
		vector<int> tmp(k);
		for (int i = 0; i <= k; i++)
		{
			if (k - i > size2)  continue;
			if (i > size1)        break;
			vector<int> num1;
			vector<int> num2;
			num1 = getnum(nums1, i);
			num2 = getnum(nums2, k - i);
			tmp = mergenum(num1, num2);
			res = cmp(res, tmp);
		}
		return res;
	}
	vector<int> getnum(vector<int>& num, int n)
	{
		vector<int> vec;  
        int drop = num.size()-n;  
        for(int i =0; i< num.size(); i++)  
        {  
            while(drop >0 && vec.size() && num[i] > vec.back())  
            {  
                drop--;  
                vec.pop_back();  
            }  
            vec.push_back(num[i]);  
        }  
        vec.resize(n);  
        return vec;  

        
	}
	vector<int> mergenum(vector<int> & num1, vector<int> & num2)
	{
		vector<int> vec;  
        while(num1.size() + num2.size() > 0)  
        {  
            vector<int>& tem = num1>num2?num1:num2;  
            vec.push_back(tem[0]);  
            tem.erase(tem.begin());  
        }  
        return vec; 


	}
	vector<int> cmp(vector<int> & num1, vector<int> & num2)
	{
		int i = 0;
		int j = 0;
		while (i < num1.size() && j < num2.size())
		{
			if (num1[i]>num2[j])
				return num1;
			if (num1[i] < num2[j])
				return num2;
				
			i++;
			j++;
		}
		return num1;
	}

};