Given an array of 2n integers, your task is to groupthese integers into n pairsof integer, say (a1, b1), (a2, b2),..., (an, bn) which makes sum of min(ai, bi) forall i from 1 to n as large as possible.
Example1:
Input: [1,4,3,2]
Output: 4
Explanation: n is 2, and the maximum sum of pairs is 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].
這個題目的意思就是給你一個偶數項的數組,然後把兩個數分一組,找出每組中最下的數,然後把這些最小的數加起來,還要讓總和儘可能的大.
其實稍微想一想就會發現,想要總和最大,就把數組排好續之後,每兩個連續的分一組,然後把每組中的第一個元素全部累加起來就好了,
實際上就是把數組中從下標爲0的元素開始的所有偶數下標都加在一起就是最大的了,如果找最小的和,就把數組排好序以後把前半部分元素加在一起就是最小的和了.
具體操作就分這兩個步驟:
1.給數組排序
2.累加偶數項和
代碼如下:
int arrayPairSum(int* nums, intnumsSize) {
int sum=0;
int i,t,p;
for (i =1; i <numsSize; ++i)
{
t=nums[i];
p=i-1;
while(p>=0 && t<nums[p])
{
nums[p+1]=nums[p];
p--;
}
nums[p+1]=t;
}
for(i=0;i<numsSize;i=i+2)
sum+= nums[i];
return sum;
}
最開始我用的冒泡排序,沒有通過,後來用了選擇排序,
