[LeetCode] 646. Maximum Length of Pair Chain

Problem:

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: 2Explanation: The longest chain is [1,2] -> [3,4]

Note:

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

解题思路：

1. 对给定的数对数组进行排序，排序的规则为：数对的第二个数较小的排在前，如[1, 2]和[2,3]由于2 < 3

    所以是[1,2]排在前，若两个数对的第二个数相等，则第一个数较小的排在前；

2. 遍历数对数组，每次找出一个最恰当的数对作为链尾，并且链的长度+1

Solution:

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

时间复杂度：O(n)