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  • (DP LIS) 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: 2
    Explanation: The longest chain is [1,2] -> [3,4]
    

    Note:

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

    ========================================================

    这个题与LIS比较相似的,在a 和 b两个数组中,我们可以只需要比较a[1]和b[0]之间的大小就行,注意要先进行排序,以子数组下标为0的标准进行排序。其他就和LIS的解法就一样了。

    C++代码:

    inline bool cmp(const vector<int> &a, const vector<int> &b){
        return a[0] < b[0];
    }
    class Solution {
    public:
        int findLongestChain(vector<vector<int>>& pairs) {
            sort(pairs.begin(),pairs.end(),cmp);
            if(pairs.size() == 0) return 0;
            int n = pairs.size();
            vector<int> dp(n,1);
            int res = 1;
            for(int i = 1; i < n; i++){
                for(int j = 0; j < i; j++){
                    if(pairs[i][0] > pairs[j][1] && dp[i] < dp[j] + 1)
                        dp[i] = dp[j] + 1;
                }
                res = max(res,dp[i]);
            }
            return res;
        }
    };
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  • 原文地址:https://www.cnblogs.com/Weixu-Liu/p/10846785.html
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