leetcode-3Sum Closest

转自：http://www.cnblogs.com/remlostime/archive/2012/10/27/2742667.html

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).

先对数组排个序。枚举第一个数，然后设两个指针，在第一个数的后半段开始王中间收缩，if sum > target则右指针往左移， if sum < target则左指针往右移。

排序O(nlogn) + 查找O(n^2) = O(n^2)

class Solution {public:    int threeSumClosest(vector<int> &num, int target) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        sort(num.begin(), num.end());                int ret;        bool first = true;                for(int i = 0; i < num.size(); i++)        {            int j = i + 1;            int k = num.size() - 1;                        while(j < k)            {                int sum = num[i] + num[j] + num[k];                if (first)                {                    ret = sum;                    first = false;                }                else                {                    if (abs(sum - target) < abs(ret - target))                        ret = sum;                                                         }                                if (ret == target)                    return ret;                                if (sum > target)                    k--;                else                    j++;            }        }                return ret;    }};