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: 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].

分析：要保证两个数之间的差最小，排序即可

class Solution {
public int arrayPairSum(int[] nums) {
quickSort(nums);
int sum = 0;
for(int i=0;i<nums.length;i=i+2){
sum += nums[i];
}
return sum;
}

private static void quickSort(int[] nums){
qsort(nums,0,nums.length-1);
}

private static void qsort(int[] nums,int left,int right){
if(left<right){
int pivot = partition(nums,left,right);
qsort(nums,left,pivot-1);
qsort(nums,pivot+1,right);
}
}

private static int partition(int[] nums,int left,int right){
int pivot = nums[left];
while(left<right){
while(left<right&&nums[right]>=pivot) --right;
nums[left] = nums[right];
while(left<right&&nums[left]<=pivot) ++left;
nums[right] = nums[left];
}
nums[left] = pivot;
return left;
}
}