leetcode 354. Russian Doll Envelopes 俄罗斯套娃 + 动态规划DP + 类似LISS最长递增子序列的做法

青旅半醒 2022-06-07 02:45 116阅读 0赞

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 (\[2,3\] => \[5,4\] => \[6,7\]).

这道题就是类似俄罗斯套娃，只需要把信封做一次排序，然后动态规划即可。

**这道题的启发就是说要学会使用DP动态规划解决问题。**

建议和[leetcode 300. Longest Increasing Subsequence 最长递增子序列LISS + 十分经典的动态规划DP做法][leetcode 300. Longest Increasing Subsequence _LISS _ _DP] 、[leetcode 368. Largest Divisible Subset 类似LISS最长递增子序列问题 + DP动态规划][leetcode 368. Largest Divisible Subset _LISS_ _ DP]和[leetcode 646. Maximum Length of Pair Chain 最长连续元素][leetcode 646. Maximum Length of Pair Chain] 一起学习

代码如下：

import java.util.ArrayList;
    import java.util.Arrays;
    import java.util.Comparator;
    import java.util.List;
    
    class Point
    {
        int width;
        int height;
        public Point(int a,int b) 
        {
            width=a;
            height=b;
        }
    }
    
    /*
     * 要习惯的使用DP去解决问题
     * dp[i]表示已i个信封结束最多有n个嵌套
     * 然后循环更新dp[i]即可
     * 这个问题挺简单的
     * */
    public class Solution 
    {
        
        public int maxEnvelopes(int[][] envelopes) 
        {
            if(envelopes==null || envelopes.length<=0)
                return 0;
    
            List<Point> list=new ArrayList<Point>();
            for(int i=0;i<envelopes.length;i++)
            {
                Point one=new Point(envelopes[i][0], envelopes[i][1]);
                list.add(one);
            }
            list.sort(new Comparator<Point>() {
                @Override
                public int compare(Point a, Point b) {
                    if(a.width!=b.width)
                        return a.width-b.width;
                    else 
                        return a.height-b.height;
                }
            });
    
            int []dp=new int[list.size()];
            Arrays.fill(dp, 1);
            for(int i=1;i<list.size();i++)
            {
                Point fa=list.get(i);
                for(int j=0;j<i;j++)
                {
                    Point now=list.get(j);
                    if(fa.height>now.height && fa.width>now.width)
                        dp[i]=Math.max(dp[i], dp[j]+1);
                }
            }
    
            int max=dp[0];
            for(int i=1;i<dp.length;i++)
                max=Math.max(max, dp[i]);
    
            return max;
        }
    }

下面是C++的做法

**仔细的想一下，本题就是LISS最长递增子序列的一个变形，所以要学会举一反三的思考和解决问题**

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <unordered_map>
    #include <set>
    #include <unordered_set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    #include <regex>
    
    using namespace std;
    
    
    
    bool cmp(pair<int, int> a, pair<int, int> b)
    {
        if (a.first != b.first)
            return a.first <= b.first;
        else
            return a.second <= b.second;
    }
    
    class Solution 
    {
    public:
        int maxEnvelopes(vector<pair<int, int>>& envelopes) 
        {
            vector<pair<int, int>> all = envelopes;
            sort(all.begin(), all.end(), cmp);
    
            int res = 0;
            vector<int> dp(all.size(), 1);
            for (int i = 0; i < all.size(); i++)
            {
                for (int j = 0; j < i; j++)
                {
                    if (all[i].first > all[j].first && all[i].second > all[j].second && dp[j] + 1 > dp[i])
                        dp[i] = dp[j] + 1;
                }
                res = max(res, dp[i]);
            }
            return res;
        }
    };

[leetcode 300. Longest Increasing Subsequence _LISS _ _DP]: http://blog.csdn.net/jackzhang_123/article/details/78122446
[leetcode 368. Largest Divisible Subset _LISS_ _ DP]: http://blog.csdn.net/jackzhang_123/article/details/78173447
[leetcode 646. Maximum Length of Pair Chain]: http://blog.csdn.net/jackzhang_123/article/details/78862409