算法--leetcode 561. Array Partition I
阿新 • • 發佈:2017-12-20
hash vector容器 一個 say 相加 整數 += blog hashtable
題目:
Given an array of 2n integers, your task is to group these integers into n pairs of integer, say (a1, b1), (a2, b2), ..., (an, bn) which makes sum of min(ai, bi) for all i from 1 to n as large as possible.
題意:
給出一個長度為2n的數組,將他們兩個一組,分為n組,求每一組中的較小值,求這些較小值相加的最大和。
輸入輸入樣例:
Input: [1,4,3,2] Output: 4 Explanation: n is 2, and the maximum sum of pairs is 4 = min(1, 2) + min(3, 4).
Note:
- n is a positive integer(正整數), which is in the range of [1, 10000].
- All the integers in the array will be in the range of [-10000, 10000].
Python 解:
思路:使用自帶函數sorted排序,將索引為0,2,4,6....n-2的數相加(即奇數順序的數),時間復雜度為nlog(n)
class Solution(object): def arrayPairSum(self, nums): """ :type nums: List[int] :rtype: int""" return sum(sorted(nums)[::2])
C++解:
思路:不懂,時間復雜度為 O(n)
語法要點:使用了vector容器,vector<int>& nums直接將 nums 數組賦值給vector容器。
vector意為向量,可以理解為數組的增強版,封裝了許多對自身操作的函數。
class Solution { public: int arrayPairSum(vector<int>& nums) { int ret = 0; bool flag = true; array<int, 20001> hashtable{ 0 }; for (const auto n : nums) { ++hashtable[n + 10000]; } for (int i = 0; i < 20001;) { if (hashtable[i] > 0) { if (flag) { flag = false; ret += (i - 10000); --hashtable[i]; } else { flag = true; --hashtable[i]; } } else ++i; } return ret; } };
算法--leetcode 561. Array Partition I