Leetcode-354. Russian Doll Envelopes

最新推荐文章于 2021-02-25 23:20:39 发布

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本文探讨了一个经典的信封套叠问题，通过动态规划的方法找到能够互相套叠的最大数量的信封组合。给出了一段C++实现代码，并通过示例说明了如何计算不同尺寸信封之间的套叠关系。

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题目：

You have a number of envelopes with widths and heights given as a pair of integers (w, h). One envelope can fit into another if and only if both the width and height of one envelope is greater than the width and height of the other envelope.

What is the maximum number of envelopes can you Russian doll? (put one inside other)

Example:
Given envelopes = [[5,4],[6,4],[6,7],[2,3]], the maximum number of envelopes you can Russian doll is 3 ([2,3] => [5,4] => [6,7]).

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源码：

class Solution {
public:
	int maxEnvelopes(vector<pair<int, int>>& envelopes) {
		int size = envelopes.size();
		if (size == 0)return 0;
		if (size == 1)return 1;
		multimap<int, pair<int, int>>info;
		int sum = 0;
		for (int i = 0; i < size; i++){
			sum = envelopes[i].first + envelopes[i].second;
			info.insert(make_pair(sum, envelopes[i]));//按和进行排序
		}
		vector<int>re(info.size(), 0);
		vector<pair<int, int>>da;
		for (map<int, pair<int, int>>::iterator i = info.begin(); i != info.end(); i++)da.push_back((*i).second);
		re[0] = 1;
		size = da.size();
		int total_max = 1;
		for (int i = 1; i < size; i++){//寻找每一个位置
			int tmp_value = 0;
			int max = 0;
			for (int j = i - 1; j >= 0; j--){
				if (i<size&&j>=0&&da[i].first>da[j].first&&da[i].second>da[j].second){
					tmp_value = re[j] + 1;
					if (tmp_value > max)max = tmp_value;
				}
			}
			if (max == 0)re[i] = 1;
			else re[i] = max;
			if (re[i] >= total_max)total_max = re[i];
		}
		return total_max;
	}
};