[Leetcode] 646. Maximum Length of Pair Chain

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

Note:

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

思路：我想到的是动态规划，但是效率不够高。用python直接超时，C++能过。dp[i]表示到第i个tuple为止最长数链长度，状态转移方程：

dp[i] = max(dp[i], pairs[i][0]>pairs[j][1]? dp[j]+1:dp[j])      j = 0,...,i-1

代码：

class Solution {public:    int findLongestChain(vector<vector<int>>& pairs) {        if(pairs.size() == 0)            return 0;                int dp[pairs.size()];        memset(dp,0,sizeof(dp));        dp[0] = 1;        sort(pairs.begin(), pairs.end(), cmp);        for(int i=1;i<pairs.size();i++){            for(int j=0;j<i;j++)                dp[i] = max(dp[i], pairs[i][0]>pairs[j][1]? dp[j]+1:dp[j]);            //  cout<<i<<"m,"<<dp[i]<<endl;        }        return dp[pairs.size()-1];    }    private:    static bool cmp(vector<int>& a, vector<int>&b) {        return a[1] < b[1];    }};

最好的方法是贪心：将数对按后一个数排序，设定两个指针cur和i。 cur指向当前数对的后一个数字，i负责往后遍历。当i碰到不满足条件的数对，直接后移；当i碰到满足条件的数对，把cur更新到i处，计数器+1。 也就是贪心地把排好序的数对，从第0个开始将符合条件的数对串起来。这样得到的链条也一定是最长的（为什么这样的贪心能保证？） 

代码：

class Solution {public:    int findLongestChain(vector<vector<int>>& pairs) {        if(pairs.size() == 0)            return 0;              int i=-1,sum=0;        sort(pairs.begin(), pairs.end(), cmp);        while(++i < pairs.size()){            sum++;            int cur = pairs[i][1];            while(i+1<pairs.size() && pairs[i+1][0] <= cur) i++;        }        return sum;    }    private:    static bool cmp(vector<int>& a, vector<int>&b) {        return a[1] < b[1];    }};