646. Maximum Length of Pair Chain思路解析

题目

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].

思路解析

要想尽量多地拼接pair，那么就应该在每次迭代的当前sub chain中末尾的pair的第二个元素值尽可能小，所以先对 Input中的pairs 按照 第二元素 升序排列，然后再遍历一次挨个判断能否连接成链条，最后链条的pair数量就是最终结果

Python实现

class Solution(object):    def findLongestChain(self, pairs):        """        :type pairs: List[List[int]]        :rtype: int        :Runtime: 112 ms        """        pairs.sort(key=operator.itemgetter(1,0))        max = -float("inf")        sum = 0        for i in pairs:            if i[0] > max:                sum += 1                max = i[1]        return sum

C++实现

// Runtime: 76 msclass Solution {private:    static bool compare(vector<int> a, vector<int> b) {        if (a[1] == b[1]) return a[0] < b[0];        return a[1] < b[1];    };    public:    int findLongestChain(vector<vector<int>>& pairs) {        std::sort(pairs.begin(), pairs.end(), compare);        int max = INT_MIN, sum = 0;        for (int i = 0; i < pairs.size(); ++i) {            if (pairs[i][0] > max) {                sum += 1;                max = pairs[i][1];            }        }        return sum;    }};

时间复杂度

不管是直接使用语言内置的sort函数还是自己实现的排序函数，都是nlogn，没有最坏情况的时间复杂度小于nlogn的算法。