561. Array Partition I

标签：pos   ati   ...   long   neu   5.0   out   ima   div   

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.

Example 1:

Input: [1,4,3,2]

Output: 4

Explanation: n is 2, and the maximum sum of pairs is 4.

Note:

1. n is a positive integer, which is in the range of [1, 10000].
2. All the integers in the array will be in the range of [-10000, 10000].

Solution 1: The sum of differences between ai and bi should be smallest and sum of the distances of adjacent elements is the smallest. The proof can be seen in https://discuss.leetcode.com/topic/87206/java-solution-sorting-and-rough-proof-of-algorithm

 1 class Solution {
 2 public:
 3     int arrayPairSum(vector<int>& nums) {
 4         sort(nums.begin(),nums.end());
 5         long long res=0;
 6         for (int i=0;i<nums.size();i+=2){
 7             res+=nums[i];
 8         }
 9         return res;
10     }
11 };

Solution 2: use bucket sort which is O(n)  https://discuss.leetcode.com/topic/87483/c-code-o-n-beats-100

561. Array Partition I

标签：pos   ati   ...   long   neu   5.0   out   ima   div   

原文地址：http://www.cnblogs.com/anghostcici/p/6916637.html