LeetCode16. 3sum closet

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).

这个题相对于3sum来说简单了许多，因为不用考虑重复的问题。基于3sum的思想，先将整个数组做一个排序。然后我们可以每次保存当前组合与target之间的差距，每次遍历找到一个新的组合时就更新这个值，每次所找到的新的组合与target之间的差距还要注意正负，如果差距为负，说明这个组合的和比target大了，就把右指针左移，否则就把左指针右移，到了最后用target减去这个gap，就能得到其和与target相距最小的组合的值。

int threeSumClosest(vector<int>& nums, int target) {
int n=nums.size();
sort(nums.begin(),nums.end());
int l=0,h=n-1;
int gap=INT_MAX;
int negative=0;
for(int i=0;i<n;i++){
int tar=target-nums[i];
l=i+1;
h=n-1;
while(l<h){
int temp=tar-nums[l]-nums[h];
if(temp==0)return target;
gap=abs(gap)<abs(temp)?gap:temp;
if(temp>0)l++;
if(temp<0)h--;
}
}
return target-gap;
}