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].

上来排序，每两个两个取就行了。但是排序得讲究方法..看了别人的代码受益匪浅

桶排序，技巧

class Solution {
public:
int arrayPairSum(vector<int>& nums) {
int a[20010] = {0};
int n = nums.size();
for (int i = 0; i < n; ++i) {
++a[nums[i] + 10000];
}
int ans = 0;
int cnt = 0;
for (int i = 0; i <= 20000;) {
if (a[i] > 0) {
cnt ++;
if (cnt&1) ans += i - 10000;
a[i] --;
}
else {
i++;
}
}
return ans;
}
};