标签:empty put tput osi 解法 ase att which sort
561. Array Partition I【easy】
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:
解法一:
1 class Solution { 2 public: 3 int arrayPairSum(vector<int>& nums) { 4 if (nums.empty()) { 5 return 0; 6 } 7 8 sort(nums.begin(), nums.end()); 9 10 int sum = 0; 11 for (int i = 0; i < nums.size(); i += 2) { 12 sum += nums[i]; 13 } 14 15 return sum; 16 } 17 };
为了不浪费元素,先排序,这样可以保证min加出来为max
比如[1, 9, 2, 4, 6, 8]
如果按顺序来的话,1和9就取1,2和4就取2,6和8就取6,显而易见并不是最大,原因就是9在和1比较的时候被浪费了,9一旦浪费就把8也给影响了,所以要先排序
@shawngao 引入了数学证明的方法,如下:
Let me try to prove the algorithm...
i
, bi >= ai
.Sm = min(a1, b1) + min(a2, b2) + ... + min(an, bn)
. The biggest Sm
is the answer of this problem. Given 1
, Sm = a1 + a2 + ... + an
.Sa = a1 + b1 + a2 + b2 + ... + an + bn
. Sa
is constant for a given input.di = |ai - bi|
. Given 1
, di = bi - ai
. Denote Sd = d1 + d2 + ... + dn
.Sa = a1 + a1 + d1 + a2 + a2 + d2 + ... + an + an + di = 2Sm + Sd
=> Sm = (Sa - Sd) / 2
. To get the max Sm
, given Sa
is constant, we need to make Sd
as small as possible.di
(distance between ai
and bi
) as small as possible. Apparently, sum of these distances of adjacent elements is the smallest. If that‘s not intuitive enough, see attached picture. Case 1 has the smallest Sd
.
标签:empty put tput osi 解法 ase att which sort
原文地址:http://www.cnblogs.com/abc-begin/p/7623481.html