3Sum Closest

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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).

分析：

先排序，然后左右夹逼，复杂度O(n^2)

//方法一:先排序，然后左右夹逼，复杂度O(n^2)
class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {

        if(nums.size() == 0)
            return 0;
        if(nums.size() == 1)
            return nums[0];
        if(nums.size() == 2)
            return nums[0] + nums[1];
        if(nums.size() == 3)
            return nums[0] + nums[1] + nums[2];

        sort(nums.begin(), nums.end());
        int result = *nums.begin() + *(nums.begin() + 1) + *(nums.end() - 1);
        for(vector<int>::iterator i = nums.begin(); i < nums.end() - 2; i++)
        {
            if(i > nums.begin() && *i == *(i-1))
                continue;
            vector<int>::iterator j = i + 1;
            vector<int>::iterator k = nums.end() - 1;

            while(j < k)
            {
                if(*i + *j + *k < target)
                {
                    if(abs(*i + *j + *k - target) < abs(result - target))
                        result = *i + *j + *k;
                    j++;
                    while(*j == *(j-1) && j < k)
                        j++;
                }
                else if(*i + *j + *k > target)
                {
                    if(abs(*i + *j + *k - target) < abs(result - target))
                        result = *i + *j + *k;
                    k--;
                    while(*k == *(k+1) && j < k)
                        k--;
                }
                else
                    return target;
            }
        }
        return result;
    }
};

//方法二：参考
class Solution {
public:
    int threeSumClosest(vector<int>& num, int target) {
        int result = 0;
        int min_gap = INT_MAX;
        sort(num.begin(), num.end());
        for (auto a = num.begin(); a != prev(num.end(), 2); ++a)
        {
            auto b = next(a);
            auto c = prev(num.end());
            while (b < c)
            {
                const int sum = *a + *b + *c;
                const int gap = abs(sum - target);
                if (gap < min_gap)
                {
                    result = sum;
                    min_gap = gap;
                }
                if (sum < target)
                    ++b;
                else
                    --c;
            }
        }
        return result;
    }
};