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

和 3 Sum 的思路接近,首先对原数组排序,随后将3 Sum 的题拆解为『1 Sum + 2 Sum』的题,对于 Closest 的题使用两根指针而不是哈希表的方法较为方便。对于有序数组来说,在查找 Cloest 的值时其实是有较大的优化空间的。

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。

复杂度分析

对原数组排序,平均时间复杂度为 O(nlogn)O(n \log n), 两重for循环,由于有两处优化,故最坏的时间复杂度才是 O(n2)O(n^2), 使用了result作为临时值保存最接近target的值,两处优化各使用了一个辅助变量,空间复杂度 O(1)O(1).

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;
    }
};

源码分析

和前面3Sum解法相似,同理使用i,j,k三个指针进行循环。
区别在于3sum中的target为0,这里新增一个变量用于比较哪组数据与target更为相近

复杂度分析

时间复杂度同理为O(n2)O(n^2) 运行时间 16ms

Reference

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