12 - 3Sum Closest

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本文介绍了一种高效算法，用于从整数数组中找到三个数，使其和最接近给定的目标值。通过两种方法对比，展示如何将时间复杂度从O(n^3)优化到O(n^2)，并提供了详细的C++实现代码。

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).

solution ; method1：O(n^3) method. 获取所有已知结果，极其低效，所以改进成第二种方法。

method2：O(n^2)方法，类似twosum方法。

method1：

class Solution {
public:
    int threeSumClosest(vector<int> &num, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int size = num.size();
        if(size < 3)
            return INT_MAX;
        
        int diff = INT_MAX;
        int result = 0;
        for(int fir = 0; fir < size -2; fir++)
        {
            int num1 = num[fir];
            
            for(int sec = fir +1; sec < size - 1; sec ++)
            {
                int num2 = num[sec];
                int thi = sec + 1;
                while(thi <= size -1)
                {
                    int num3 = num[thi];
                    int sum = num1 + num2 + num3;
                    int tempDiff = abs(sum - target);
                    if(tempDiff < diff)
                    {
                        result = sum;
                        diff = tempDiff;
                    }
                        
                    thi ++;
                }
            }
        }
        return result;
    }
};

method2:

class Solution {
public:
    int threeSumClosest(vector<int> &num, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int size = num.size();
        if(size < 3)
            return INT_MAX;
        
        int diff = INT_MAX;
        int result = 0;
        sort(num.begin(),num.end());
        for(int fir = 0; fir < size -2; fir++)
        {
            int num1 = num[fir];
            
            int start = fir + 1;
            int end = size -1;
            while(start < end)
            {
                int sum = num1 + num[start] + num[end];
                int tempDiff = abs(sum - target);
                if(tempDiff < diff)
                {
                    diff = tempDiff;
                    result = sum;
                }
                
                if(sum == target)
                    return sum;
                else if(sum > target)
                    end--;
                else
                    start++;
            }
        }
        return result;
    }
};