【一天一道LeetCode】#16. 3Sum Closest

一天一道LeetCode系列

（一）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).

（二）解题

直接用三重循环，然后考虑到重复的数字，则需要先排序，以便于后续去重。
其次，当等于target时，则直接返回

class Solution {public:    int threeSumClosest(vector<int>& nums, int target) {        int min = 2147438647;        int key =0;        std::sort(nums.begin() , nums.end());        for(int i = 0  ; i < nums.size()-2 ; )        {            for(int j = i+1 ; j < nums.size()-1 ;)            {                for(int k = j+1 ; k < nums.size() ; )                {                    int gap = nums[i]+nums[j]+nums[k];                    int temp = gap-target>0?gap-target:target-gap;                    if(temp<min){                         min = temp;                        key = gap;                        if(min ==0)  //如果找到等于0的则返回                        {                            return key;                        }                    }                    k++;                    while(k<nums.size() && nums[k] == nums[k-1]) ++k;                }                j++;                while(j<nums.size()-1 && nums[j] == nums[j-1]) ++j;            }            i++;            while(i<nums.size()-2 && nums[i] == nums[i-1]) ++i;        }        return key;    }};

在网上看到另外一种快速的解法。O（n^2）

class Solution {public:    int threeSumClosest(vector<int>& nums, int target) {        std::sort(nums.begin() , nums.end());        bool isfirst = true;        int ret;        for(int i = 0  ; i < nums.size() ; i++)        {            int j = i+1;            int k = nums.size()-1;            while(j<k){                int sum = nums[i]+nums[j]+nums[k];                if(isfirst)                {                    ret = sum;                    isfirst = false;                }                else                {                    if(abs(sum - target) < abs(ret - target))                    {                        ret = sum;                    }                }                if(ret == target)                    return ret;                if(sum>target)                    k--;                else                    j++;            }        }        return ret;    }};