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  • 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].

    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];
        }
    }
    

      

    永远渴望,大智若愚(stay hungry, stay foolish)
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  • 原文地址:https://www.cnblogs.com/h-hkai/p/10511175.html
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