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].
上来排序,每两个两个取就行了。但是排序得讲究方法..看了别人的代码受益匪浅
桶排序,技巧
class Solution { public: int arrayPairSum(vector<int>& nums) { int a[20010] = {0}; int n = nums.size(); for (int i = 0; i < n; ++i) { ++a[nums[i] + 10000]; } int ans = 0; int cnt = 0; for (int i = 0; i <= 20000;) { if (a[i] > 0) { cnt ++; if (cnt&1) ans += i - 10000; a[i] --; } else { i++; } } return ans; } };
原文地址:http://www.cnblogs.com/pk28/
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