016 最接近的三數之和

LeetCode 第十六題 最接近的三數之和

給定一個包括 n 個整數的數組 S，找出 S 中的三個整數使得他們的和與給定的數 target 最接近。返回這三個數的和。假定每組輸入只存在一個答案。
例如，給定數組 S = {-1 2 1 -4}, 並且 target = 1.

與 target 最接近的三個數的和爲 2. (-1 + 2 + 1 = 2).

其實這道題和第十五題基本類似，可以參考十五題的解題步驟。

Java

    public static int threeSumClosest(int[] nums, int target) {
        int result = 0;
        if (nums.length <= 3) {
            for (int i : nums)
                result += i;
            return result;
        }
        Arrays.sort(nums);
        result = nums[0] + nums[1] + nums[2];
        for (int i = 0; i < nums.length - 2; i++) {
            int low = i + 1, high = nums.length - 1;
            while (low < high) {
                int current_sum = nums[i] + nums[low] + nums[high];
                if (Math.abs(target - current_sum) < Math.abs(target - result)) {
                    result = current_sum;
                    if (result == target)
                        return target;
                }
                if (current_sum > target)
                    high--;
                else
                    low++;
            }
        }
        return result;
    }

Python

class Solution(object):
    def threeSumClosest(self, nums, target):
        if len(nums) <= 3:
            return sum(nums)
        nums = sorted(nums)
        result = nums[0] + nums[1] + nums[2]
        for i in range(len(nums) - 2):
            low, high = i + 1, len(nums) - 1
            while (low < high):
                current_sum = nums[i] + nums[low] + nums[high]
                if (abs(target - current_sum) < abs(target - result)):
                    result = current_sum
                    if (result == target):
                        return result
                if (current_sum > target):
                    high = high - 1
                else:
                    low = low + 1
        return result

C++

class Solution
{
public:
    int threeSumClosest(vector<int>& nums, int target)
    {
        int result=0;
        if(nums.size()<=3)
        {
            for(int i=0; i<nums.size(); i++)
            {
                result+=nums[i];
            }
            return result;
        }
        sort(nums.begin(),nums.end());
        result=nums[0]+nums[1]+nums[2];
        for(int i=0; i<nums.size()-2; i++)
        {
            int low=i+1,high=nums.size()-1;
            while(low<high)
            {
                int current_sum=nums[i]+nums[low]+nums[high];
                if(abs(target-current_sum)<abs(target-result))
                {
                    result=current_sum;
                    if(result==target)return result;
                }
                if(current_sum>target)
                    high--;
                else
                    low++;
            }
        }
        return result;
    }
};