[LeetCode] 3Sum 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).

分析：和3Sum一样，不能用O（n^3）遍历，需要有一些可以简化的步骤，三层遍历肯定Time Limited Exceeded！

class Solution {
public:
int threeSumClosest(vector<int> &num, int target) {
int n = num.size();
map<int,int> diff;//key是target和sum的差值的绝对值，value是sum值
sort(num.begin(),num.end());
int sum ;
for(int i=1;i<n-1;i++){

int l = 0, r= n-1;
while(l<i && r>i){
if(l>1 && num[l] == num[l-1]){
l++;
continue;
}
if(r<n-1 && num[r] == num[r+1]){
r--;
continue;
}
sum = num[l]+num[i]+num[r];
if(sum == target)
return sum;
else {
int abs   = sum-target>0 ? sum-target :target-sum;
diff[abs] = sum;
if(sum >target)
r--;
else
l++;

}
}//end while
}//end for
map<int,int>::iterator iter = diff.begin();
return (*iter).second;
}
};