LeectCode 16. 3Sum Closest *** 对撞指针

题目

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).

题意

给定一个含有n个元素的整型数组S，从S中找出3个元素使得其和与target最接近，返回这三个元素的和

注意

会有无解的情况吗?
有多解，就是其和与target值得接近程度一样，怎么办？这些都应该和面试管沟通清楚。该题已经说明有唯一解。

思路

类似15.3sum。
a+b+c = sum，遍历c，对撞a，b每次计算sum和target的接近程度。
重点是:如何判断两个数的接近程度，在数轴上表示就是两点之间的距离
x1和x2之间的距离 distance = abs(x1-x2)，当前的距离比历史的距离更小则更新并记录

代码

class Solution {public:    int threeSumClosest(vector<int>& nums, int target) {        int min=9999;        int minsum = 0;        sort(nums.begin(),nums.end());        //排序，对撞的必要条件        for(int i=0;i<nums.size();i++)        {            int l=i+1;            int r=nums.size()-1;            while(l<r)            {                int tempmin = 0;                //动态调整不断逼近target                if(target>(nums[l]+nums[r]+nums[i]))                {                    tempmin = nums[i]+nums[l]+nums[r];                    l++;                }                else                {                    tempmin = nums[i]+nums[l]+nums[r];                    r--;                }               //定义三个元素的和与目标的接近程度                if(abs(target-tempmin)<min)               {                   min = abs(target - tempmin);                   minsum = tempmin;               }            }        }        return minsum;    }};

结果