Leetcode_3sum-closest

地址：http://oj.leetcode.com/problems/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).

思路：上一题的变形。参考：http://blog.csdn.net/flyupliu/article/details/24438255
参考代码：

class Solution {
public:
int threeSumClosest(vector<int> &num, int target) {
int min_val = 1<<30, ans = 0;
sort(num.begin(), num.end());
for(int i = 0; i < num.size()-2; ++i)
{
if(i && num[i]==num[i-1])
continue;
int left = i + 1, right = num.size() - 1;
while(left < right)
{
if(right<num.size()-2 && num[right]==num[right+1])
{
--right;
continue;
}
if(num[i]+num[left]+num[right]==target)
return target;
else if (num[i]+num[left]+num[right]>target)
{
if(min_val > num[i]+num[left]+num[right] - target)
{
ans = num[i]+num[left]+num[right];
min_val = num[i]+num[left]+num[right] - target;
}
--right;
}
else
{
if(min_val > target - (num[i]+num[left]+num[right]))
{
ans = num[i]+num[left]+num[right];
min_val = target-(num[i]+num[left]+num[right]);
}
++left;
}
}
}
return ans;
}
};

SECOND TRIAL

class Solution {
public:
int threeSumClosest(vector<int> &num, int target) {
int diff = INT_MAX, ans = 0;
sort(num.begin(), num.end());
for(int i = 0; i<num.size()-2; ++i)
{
if(i && num[i]==num[i-1])
continue;
for(int j = num.size()-1; j>i+1; --j)
{
if(j < num.size()-1 && num[j]==num[j+1])
continue;
for(int k = i + 1; k < j; ++k)
{
if(k>i+1 && num[k]==num[k-1])
continue;
if(abs(num[i]+num[j]+num[k]-target)<diff)
{
diff = abs(num[i]+num[j]+num[k]-target);
ans = num[i]+num[j]+num[k];
if(!diff)
return target;
}
}
}
}
return ans;
}
};