646. Maximum Length of Pair Chain

Posted 2021-02-12 ruruozhenhao

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].

 

Approach #1: Greedy. [C++]

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

　　

 

Approach #2: DP. [Java]

class Solution {
    public int findLongestChain(int[][] pairs) {
        if (pairs == null || pairs.length == 0) return 0;
        Arrays.sort(pairs, (a, b) -> (a[0] - b[0]));
        int[] dp = new int[pairs.length];
        Arrays.fill(dp, 1);
        for (int i = 0; i < dp.length; ++i) {
            for (int j = 0; j < i; ++j) {
                dp[i] = Math.max(dp[i], pairs[i][0] > pairs[j][1] ? dp[j] + 1: dp[j]);
            }
        }
        return dp[pairs.length - 1];
    }
}