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 = 0, right = 0, val = 0, minC = INT_MAX;

    sort(num.begin(), num.end());

    for (int i = 0; i < num.size() - 2; ++i)

    {

        left = i + 1; right = num.size() - 1;

        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 > 0)

                left++;

            else if (tepVal < 0)

                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 = 0; i < num.size(); ++i){

            if (i > 0 && num[i] == num[i-1])

                continue;

            int left = i+1, right = num.size()-1;

            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 == 0)

                    return target;

                else if(val > 0)

                    left++;

                else

                    right--;

            }

        }

        return ans;

    }

};