LeetCode 561: Array Split I Array Partition I

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算法是一个程序的灵魂。
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.

A given length of 2n array, your task is divided into the number of these n pairs, such as (a1, b1), (a2 , b2), ..., (an, bn), such that from 1 to n min ( ai, bi) the sum of the maximum.

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:

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].

prompt:

1. n is a positive integer in the range [1, 10000].
2. Range of elements in the array [-10000, 10000].

Problem-solving ideas:

In fact, the array is sorted, and then sequentially each of the two numbers is both a pair, each pair of the first number and the accumulated sum is also desired. It is about to test the performance of various types of sorting algorithms.

First use the built-in sort()function to understand some ideas:

Java：

import java.util.Arrays;
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;
    }
}

Extended:

Wiki encyclopedia of sorting algorithms described in great detail, and are classified compared address: https://zh.wikipedia.org/wiki/%E6%8E%92%E5%BA%8F%E7%AE%97 % E6% B3% 95

Here two simple recommendation:

• Quick sort (quick sort) - the expected time, the worst case; for large, random number list is generally believed to be the fastest known sorting (C language standard library qsort()for sorting is quick sort algorithm, recursive thinking and divide and rule)
• Bucket sort (the Sort bucket) - ; require additional space (typical sacrifice space for time)

Bubble sort, selection sort are relatively simple and easy to understand the complexity of Shi n^2, so will not repeat them.