leetcode第16题--3Sum Closest

Problem：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).

吃饭之前刷了这题。思路就是和上题类似。先srot，然后固定i从第一个到倒数第三个，然后设一个left=i+1，right=num.size()-1; 如果三个数相加和target相比较大，那就right--，使三个数相加要变小。如果比target小那就left++。同时用abs记录当前的差值，如果比之前的差值更小，那就记录三个数的和到val中。如果差值为零了，那就直接返回三个数的和。否则到最后在返回val就可以了。代码如下

class Solution {
public:
int threeSumClosest(vector<int> &num, int target)
{
    int left = , right = , val = , minC = INT_MAX;
    sort(num.begin(), num.end());
    for (int i = ; i < num.size() - ; ++i)
    {
        left = i + ; right = num.size() - ;
        while(left < right)
        {
            int tepVal = target - num[i] - num[left] - num[right];
            if (abs(tepVal) < minC)
                {minC = abs(tepVal); val = num[i] + num[left] + num[right];}
            if (tepVal > )
                left++;
            else if (tepVal < )
                right--;
            else
                return num[i] + num[left] + num[right];
        }
    }
    return val;
}
};

吃饭去了。

2015/4/3:

class Solution {
public:
    int threeSumClosest(vector<int> &num, int target) {
        sort(num.begin(), num.end());
        int ans, globalVal = INT_MAX;
        for (int i = ; i < num.size(); ++i){
            if (i >  && num[i] == num[i-])
                continue;
            int left = i+, right = num.size()-;
            while(left < right){
                int val =target - (num[i] + num[left] + num[right]);
                if (abs(val) < globalVal){
                    ans = num[i] + num[left] + num[right];
                    globalVal = abs(val);
                }
                if (val == )
                    return target;
                else if(val > )
                    left++;
                else
                    right--;
            }
        }
        return ans;
    }
};