# 3 Sum Closest

## Question

Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target.
Return the sum of the three integers. You may assume that each input would have exactly one solution.

For example, given array S = {-1 2 1 -4}, and target = 1.
The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).


## 题解1 - 排序 + 2 Sum + 两根指针 + 优化过滤

### Python

class Solution:
"""
@param numbers: Give an array numbers of n integer
@param target : An integer
@return : return the sum of the three integers, the sum closest target.
"""
def threeSumClosest(self, numbers, target):
result = 2**31 - 1
length = len(numbers)
if length < 3:
return result

numbers.sort()
larger_count = 0
for i, item_i in enumerate(numbers):
start = i + 1
end = length - 1
# optimization 1 - filter the smallest sum greater then target
if start < end:
sum3_smallest = numbers[start] + numbers[start + 1] + item_i
if sum3_smallest > target:
larger_count += 1
if larger_count > 1:
return result

while (start < end):
sum3 = numbers[start] + numbers[end] + item_i
if abs(sum3 - target) < abs(result - target):
result = sum3

# optimization 2 - filter the sum3 closest to target
sum_flag = 0
if sum3 > target:
end -= 1
if sum_flag == -1:
break
sum_flag = 1
elif sum3 < target:
start += 1
if sum_flag == 1:
break
sum_flag = -1
else:
return result

return result


### 源码分析

1. leetcode 上不让自己导入sys包，保险起见就初始化了result为还算较大的数，作为异常的返回值。
2. 对数组进行排序。
3. 依次遍历排序后的数组，取出一个元素item_i后即转化为『2 Sum Cloest』问题。『2 Sum Cloest』的起始元素索引为i + 1，之前的元素不能参与其中。
4. 优化一——由于已经对原数组排序，故遍历原数组时比较最小的三个元素和target值，若第二次大于target果断就此罢休，后面的值肯定越来越大。
5. 两根指针求『2 Sum Cloest』，比较sum3resulttarget的差值的绝对值，更新result为较小的绝对值。
6. 再度对『2 Sum Cloest』进行优化，仍然利用有序数组的特点，若处于『一大一小』的临界值时就可以马上退出了，后面的元素与target之差的绝对值只会越来越大。

### C++

class Solution {
public:
int threeSumClosest(vector<int> &num, int target)
{
if (num.size() <= 3) return accumulate(num.begin(), num.end(), 0);
sort (num.begin(), num.end());

int result = 0, n = num.size(), temp;
result = num[0] + num[1] + num[2];
for (int i = 0; i < n - 2; ++i)
{
int j = i + 1, k = n - 1;
while (j < k)
{
temp = num[i] + num[j] + num[k];

if (abs(target - result) > abs(target - temp))
result = temp;
if (result == target)
return result;
( temp > target ) ? --k : ++j;
}
}
return result;
}
};