646. Maximum Length of Pair Chain

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本文介绍了一个寻找最长数对链的问题，通过排序而非动态规划的方法高效解决。给定一系列递增数对，目标是找到能够形成最长连续递增序列的数对组合。

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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].

这道题目不用dp，采用排序即可完成。

class Solution {
    public:
        int findLongestChain(vector<vector<int>>& pairs) {
            sort(pairs.begin(), pairs.end(), cmp);
            int cnt = 0;
            vector<int>& pair = pairs[0];
            for (int i = 0; i < pairs.size(); i++) {
                if (i == 0 || pairs[i][0] > pair[1]) {
                    pair = pairs[i];
                    cnt++;
                }
            }
            return cnt;
        }

    private:
        static bool cmp(vector<int>& a, vector<int>&b) {
            return a[1] < b[1] || a[1] == b[1] && a[0] < b[0];
        }/*按照a[0]的大小来排序*/
};