leetcode Question 2: 3 sum closest

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).

The code is slight different from the 3 sum problem, just change the if condition, the key point is to use the abs() function to get the distances between the target and the output value.

The source code:

class Solution {
public:
int threeSumClosest(vector<int> &num, int target) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int res;
vector<int>::iterator it1,it2,it3,iend;

sort(num.begin(),num.end());
if(num.size()>=3){
res = num.at(0)+num.at(1)+num.at(2);
iend = num.end();
it1 = num.begin();
bool f1 = false;
for ( ;it1!=iend-2;++it1){
if(f1 && *it1==*(it1-1)) {continue;}
f1 = true;
bool f2 = false;
bool f3 = false;
it3 = iend-1;
it2 = it1+1;
while (it2!=it3){

if(f3 && *it3==*(it3+1)) { it3 = it3 -1 ;continue;}
if(f2 && *it2==*(it2-1)) { it2 = it2 +1 ;continue;}

if (*it1 + *it2 + *it3==target){
res = target;
return res;
}
if (*it1 + *it2 + *it3 >target){
if (abs(target - (*it1 + *it2 + *it3)) < abs(target-res)){
res = *it1 + *it2 + *it3;
}
it3=it3-1;
f3 = true;

}else{
if (abs(target - (*it1 + *it2 + *it3)) < abs(target-res)){
res = *it1 + *it2 + *it3;
}
it2=it2+1;
f2 = true;
}

}

}
}
return res;
}
};


1 comment:

1. Hi, can we reduce the lines of code anyhow? I mean can we think of a shorter and efficient code like this?