Array_Partition_I.md

561. Array Partition I

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 = min(1, 2) + min(3, 4).

Note:

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

Method:

排序后将数组下标为偶数的元素求和即是，时间复杂度O(nlogn):

class Solution(object):
    def arrayPairSum(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        nums.sort()
        ans=0
        for i in range(0,len(nums)-1,2):
            ans+=nums[i]
        return ans

or:

        nums.sort()
        return sum(nums[i] for i in range(0, len(nums)-1, 2))

counting sort when elements are in certain ranges:

class Solution(object):
    def arrayPairSum(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        cnt=[0]*20001
        for i in range(len(nums)):
            cnt[nums[i]+10000]+=1
        res, flag=0, 0
        i=0
        while i<len(cnt):
            if cnt[i]>0 and flag==0:
                res+=i-10000
                cnt[i]-=1
                flag=1
            elif cnt[i]>0 and flag==1:
                cnt[i]-=1
                flag=0
            else:
                i+=1
        return res