LeetCode 561. Array Partition I(数组划分 I)
题目描述
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].
解法一
思路
这题看起来有点绕,理清楚了就会发现很简单了。
首先要明确一点,最后是需要在2n个数里面找出n个(不重复)的数,即每个数用一次。再来看“结果最大”,所以我们希望找到的n个数尽可能大。那么假设2n个数是有序的,考虑最后一个数,这个数最大,但却不可能在结果中,因为min(a, b)
永远都取不到最大的数。那再来看看倒数第二大的数,为了让这个数能够在结果中,我们就只能找一个比它更大的数来匹配,而显然这个数只能是最后那个最大的数。顺着这个思路,就会发现要取到的数一定是数组排序后,偶数index的数,最后全部相加就是最终结果啦。
Python
class Solution(object):
def arrayPairSum(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
nums.sort()
return sum(nums[::2])
Java
public class Solution {
public int arrayPairSum(int[] nums) {
Arrays.sort(nums);
int sum = 0;
for (int i = 0; i < nums.length; i += 2) {
sum += nums[i];
}
return sum;
}
}
C++
class Solution {
public:
int arrayPairSum(vector<int> &nums) {
sort(nums.begin(), nums.end());
int sum = 0;
for (int i = 0; i < nums.size(); i += 2) {
sum += nums[i];
}
return sum;
}
};