646. Maximum Length of Pair Chain思路解析

題目

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

要想盡量多地拼接pair，那麼就應該在每次迭代的當前sub chain中末尾的pair的第二個元素值儘可能小，所以先對 Input中的pairs 按照 第二元素 升序排列，然後再遍歷一次挨個判斷能否連接成鏈條，最後鏈條的pair數量就是最終結果

Python實現

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++實現

// Runtime: 76 ms

class 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;
    }
};

時間複雜度

不管是直接使用語言內置的sort函數還是自己實現的排序函數，都是nlogn，沒有最壞情況的時間複雜度小於nlogn的算法。