[leetcode-646-Maximum Length of Pair Chain]-优快云博客

本文介绍了一种使用动态规划解决寻找最长数对链的问题。通过定义递推公式 dp[i]=max(dp[i],dp[j]+1)，并在一系列已排序的数对中找到满足条件的最长链。

You are given n pairs of numbers. In every pair, the first number is always smaller than the second number.

Now, we define a pair (c, d) can follow another pair (a, b) if and only if b < c. Chain of pairs can be formed in this fashion.

Given a set of pairs, find the length longest chain which can be formed. You needn't use up all the given pairs. You can select pairs in any order.

Example 1:

Input: [[1,2], [2,3], [3,4]]
Output: 2
Explanation: The longest chain is [1,2] -> [3,4]

Note:

The number of given pairs will be in the range [1, 1000].

思路：
动态规划。递推公式是 dp[i] = max(dp[i],dp[j]+1);其中0=<j<i;

int findLongestChain(vector<vector<int>>& pairs)
    {
        sort(pairs.begin(), pairs.end(), [](const vector<int>&a, const vector<int>&b){return a[0] < b[0]; });
         
        vector<int> dp(pairs.size(), 1);
        for (int i = 1; i < pairs.size(); i++)
        {
            for (int j = 0; j<i;j++)
            {
                if (pairs[i][0]>pairs[j][1])
                {
                    dp[i] = max(dp[i],dp[j]+1);                    
                }
            }
            
        }
        return *max_element(dp.begin(), dp.end());
    }