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].思路：
我们可以利用类似贪心算法分析此问题，若想子链越长，则每个节点的第二位元素尽可能小，所以我们维护一个最小堆，将数组集按数组的第二位元素从小到大排序，然后遍历每个节点，若符合链状结构，则将长度加1
代码1：
class Solution { public: int findLongestChain(vector<vector<int>>& pairs) { sort(pairs.begin(), pairs.end(), [](vector<int> f, vector<int> s){return f[1] < s[1];}); int res = 1, last = 0; for (int i = 0; i < pairs.size(); i++) { if (pairs[i][0] > pairs[last][1]) { last = i; res++; } } return res; } };