16. 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).

class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
int res = target;
std::sort(nums.begin(),nums.end());
int ans = nums[0] + nums[1] + nums[2];
for(int i = 0; i < nums.size(); ++i){
int front = i+1, back = nums.size()-1;
while(front < back){
if(nums[i] + nums[front] + nums[back] < res){
if(res - (nums[i] + nums[front] + nums[back]) < abs(ans - res))
ans = (nums[i] + nums[front] + nums[back]);
front++;

}
else if(nums[i] + nums[front] + nums[back] > res){
if((nums[i] + nums[front] + nums[back]) - res < abs(ans - res))
ans = (nums[i] + nums[front] + nums[back]);
back--;
}
else
return res;

while(front < back && nums[front] == nums[front+1]) // 错写成nums[front] == nums[front++].... sha
front++;
while(front < back && nums[back] == nums[back-1])
back--;
}

while(i+1 < nums.size() && nums[i] == nums[i+1])
++i;
}
return ans;

}
};