[leetcode]354. Russian Doll Envelopes 俄罗斯套娃信封问题

原创已于 2024-07-10 18:06:21 修改 · 337 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

于 2017-07-09 10:55:08 首次发布

leetcode 同时被 2 个专栏收录

473 篇文章

订阅专栏

动态规划

91 篇文章

订阅专栏

本文探讨了LeetCode上的俄罗斯套娃信封问题，通过给出的一段C++代码实现了解决方案。该问题要求找到能相互套叠的最大数量的信封集合。采用排序和动态规划的方法来解决这一挑战。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目链接：https://leetcode.com/problems/russian-doll-envelopes/#/description

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]).

bool cmp (pair<int, int> i, pair<int, int> j) {
    if (i.first == j.first)
        return i.second < j.second;
    return i.first < j.first;
}

class Solution {
public:
    int maxEnvelopes(vector<pair<int, int>>& envelopes) {
        int N = envelopes.size();
        vector<int> dp(N, 1);
        int mx = (N == 0) ? 0 : 1;
        sort(envelopes.begin(), envelopes.end(), cmp);
        for (int i = 0; i < N; i++) {
            for (int j = i - 1; j >= 0; j--) {
                if (envelopes[i].first > envelopes[j].first && envelopes[i].second > envelopes[j].second) {
                    dp[i] = max(dp[i], dp[j] + 1);
                    mx = max(dp[i], mx);
                }
            }
        }
        return mx;
    }
};