561. Array Partition

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Problem

Given an integer array nums of 2n integers, group these integers into n pairs (a1, b1), (a2, b2), …, (an, bn) such that the sum of min(ai, bi) for all i is maximized. Return the maximized sum.

Example 1:

Input: nums = [1,4,3,2]
Output: 4
Explanation: All possible pairings (ignoring the ordering of elements) are:
1. (1, 4), (2, 3) -> min(1, 4) + min(2, 3) = 1 + 2 = 3
2. (1, 3), (2, 4) -> min(1, 3) + min(2, 4) = 1 + 2 = 3
3. (1, 2), (3, 4) -> min(1, 2) + min(3, 4) = 1 + 3 = 4
   So the maximum possible sum is 4.

Example 2:

Input: nums = [6,2,6,5,1,2]
Output: 9
Explanation: The optimal pairing is (2, 1), (2, 5), (6, 6). min(2, 1) + min(2, 5) + min(6, 6) = 1 + 2 + 6 = 9.

Constraints:

• $1 <= n <= 10^4$
• nums.length == 2 * n
• $-10^4 <= nums[i] <= 10^4$

Code

class Solution {
    public int arrayPairSum(int[] nums) {
        Arrays.sort(nums);
        
        int res = 0;
        for (int i = 0; i < nums.length;  i++) {
            if (i % 2 == 0) {
                res += nums[i];
            }
        }
        
        return res;
    }
}

class Solution {
    public int arrayPairSum(int[] nums) {
        int range = 10000;
        int[] count = new int[2 * range + 1];
        
        for (int num : nums) {
            count[num + range]++;
        }
        
        int res = 0;
        boolean isEvenIndex = true;
        
        for (int i = 0; i <= 2 * range; i++) {
            while (count[i] > 0) {
                if(isEvenIndex) {
                    res += i - range;
                }

                isEvenIndex = !isEvenIndex;

                count[i]--;
            }
        }
        
        return res;
    }
}