LeetCode 561. Array Partition I

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最新推荐文章于 2022-01-31 22:01:47 发布

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本文介绍了一种通过排序和选择每个偶数位置上的元素来最大化整数对最小值之和的方法。示例输入为[1,4,3,2]时，输出为4，即(min(1,2)+min(3,4))。使用快速排序算法进行排序，适用于范围在[1,10000]内的正整数n，以及[-10000,10000]内的整数。

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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), ..., (a_n, b_n) which makes sum of min(a_i, b_i) 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].

Java：

class Solution {
    public int arrayPairSum(int[] nums) {
        int sum = 0;
        quickSort(nums,0,nums.length - 1);
        for (int i = 0; i < nums.length; i += 2) {
            sum += nums[i];
        }
        return sum;
    }
     public void quickSort(int[] array, int low, int high) {
            if (low > high) return;
            int i = low;
            int j = high;
            int temp = array[low];
            while (i != j) {
                while (array[j] >= temp && i < j) {
                    j--;
                }
                while (array[i] <= temp && i < j) {
                    i++;
                }
                if (array[i] > array[j]) {
                    int tem = array[i];
                    array[i] = array[j];
                    array[j] = tem;
                }
            }
            array[low] = array[i];
            array[i] = temp;
            quickSort(array,low,j - 1);
            quickSort(array,i + 1,high);
            return;
        }
}