LeetCode-561 Array Partition I

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: 4Explanation: 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].

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根据题意我的思路是对数组进行排序，然后取偶数位相加即可：

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

这种方法思路简单，但是用时较长，估计用在排序上面，38ms，如果对时间限制比较苛刻的话就不行了。

还有一种方法，思路和我的差不多，但是实现方法却比我好很多，耗时14ms:

public class Solution {    public int arrayPairSum(int[] nums) {        if(nums == null || nums.length == 0) return 0;//carry用来表示奇偶        int sum = 0, carry = 0;//Note（2）：integers will be in the range of [-10000,10000].         int[] map = new int[20001];// 作用相似于对数组进行了排序，        for(int i : nums) map[i + 10000]++;        for(int i = 0; i < map.length; i++) {            if(map[i] == 0) continue;    //carry==0为偶数时，额外加1实现对偶数位求和            sum += (map[i] - carry + 1) / 2 * (i - 10000);    //偶数位为0，奇数位为1            carry = (map[i] - carry) % 2;        }        return sum;    }}

这个方法新颖的地方有两点：

1是第一个for循环。将源数组的值加10000作为新数组的索引，然后利用索引进行计算，比Arrays.sort()快很多

2是求和时的计算。

下面是运行过程：

carry: 0 |(map[i]-carry +1): 1 |i-10000: 1 |sum: 1

carry: 1 |(map[i]-carry +1): 0 |i-10000: 3 |sum: 1

carry: 0 |(map[i]-carry +1): 1 |i-10000: 5 |sum: 6

carry: 1 |(map[i]-carry +1): 0 |i-10000: 6 |sum: 6

carry: 0 |(map[i]-carry +1): 1 |i-10000: 7 |sum: 13

carry: 1 |(map[i]-carry +1): 0 |i-10000: 8 |sum: 13