[LeetCode] Array Partition I 數組分割之一

 

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.

 

Note:

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

 

這道題讓我們分割數組，兩兩一對，讓每對中較小的數的和最大。這題難度不大，用貪婪算法就可以了。由於我們要最大化每對中的較小值之和，那么肯定是每對中兩個數字大小越接近越好，因為如果差距過大，而我們只取較小的數字，那么大數字就浪費掉了。明白了這一點，我們只需要給數組排個序，然后按順序的每兩個就是一對，我們取出每對中的第一個數即為較小值累加起來即可，參見代碼如下：

 

class Solution {
public:
    int arrayPairSum(vector<int>& nums) {
        int res = 0, n = nums.size();
        sort(nums.begin(), nums.end());
        for (int i = 0; i < n; i += 2) {
            res += nums[i];
        }
        return res;
    }
};

 

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