地址：
https://leetcode.com/problems/array-partition-i/
题目：
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].
题目思路
- 两两分组，找出每组最小值加起来最大的解决方案，如果一大一小两个一组，大的就浪费了，所以按照大小排序，相邻的一组，故加起来是偶数序号的数字之和
- 解题关键是数组排序，我自己的做法是用的java自己的Arrays.sort(）方法
解法1（自己的解法）
public static int arrayPairSum(int[] nums) {
Arrays.sort(nums);
int sum=0;
for(int i=0;i<nums.length;i+=2){
sum+=nums[i];
}
return sum;
}
26 ms, faster than 53.83%
解法2
public class Solution {
public int arrayPairSum(int[] nums) {
int[] exist = new int[20001];
for (int i = 0; i < nums.length; i++) {
exist[nums[i] + 10000]++;
}
int sum = 0;
boolean odd = true;
for (int i = 0; i < exist.length; i++) {
while (exist[i] > 0) {
if (odd) {
sum += i - 10000;
}
odd = !odd;
exist[i]--;
}
}
return sum;
}
}
这个 O(n) beats 100%确实厉害，运用了桶排序的思路。
接下来学习排序算法。
