561. Array Partition I
2017-09-20 09:57
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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:
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].
题目的要求是将所给数组划分成n对,取出每对整数中的最小值,依次相加求和,通过划分使得求得的和最大。
通过例子,我们不难发现,其实最佳的划分思路就是按照大小排序之后去划分,其实这种划分思路说到低用的是贪心的思想,让在相同位置上每次取得的min都尽可能得大,然后求和。这样代码的思路就十分清晰了:先排序,再求和:
可以看到,我是先对数组排序,然后按照可以划分的对数,算出每一对的最小值,依次相加。事实上,在排完顺序之后,完全可以直接求和。因为我们是每一对去求最小值,在排完序之后,我们直接间隔取值就可以了:
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].
题目的要求是将所给数组划分成n对,取出每对整数中的最小值,依次相加求和,通过划分使得求得的和最大。
通过例子,我们不难发现,其实最佳的划分思路就是按照大小排序之后去划分,其实这种划分思路说到低用的是贪心的思想,让在相同位置上每次取得的min都尽可能得大,然后求和。这样代码的思路就十分清晰了:先排序,再求和:
class Solution { public: int arrayPairSum(vector<int>& nums) { int size = nums.size() / 2; sort(nums.begin(), nums.end()); int sum = 0; for (int i = 0 ; i < size; i++) { int min = 1000000; for (int j = 2 * i; j <= 2 * i + 1; j++) { if (nums[j] < min) min = nums[j]; } sum += min; } return sum; } };
可以看到,我是先对数组排序,然后按照可以划分的对数,算出每一对的最小值,依次相加。事实上,在排完顺序之后,完全可以直接求和。因为我们是每一对去求最小值,在排完序之后,我们直接间隔取值就可以了:
class Solution { public: int arrayPairSum(vector<int>& nums) { //int size = nums.size() / 2; sort(nums.begin(), nums.end()); int sum = 0; for (int i = 0; i < nums.size(); i += 2) { sum += nums[i]; } return sum; } };
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