leetcode 57: 3Sum Closest

3Sum ClosestJan
18 '12

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> &num, int target) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(num.size()<3) return 0;

int min=num[0]+num[1]+num[2];
sort(num.begin(), num.end() );
int sz=num.size();
int j,k;
for(int i=0; i<sz-2;i++) {
j=i+1;
k=sz-1;
while(j<k) {
int sum = num[i]+num[j]+num[k];
int temp = sum-target;
if( temp == 0 ) return target;
else if( temp < 0 ) j++;
else k--;

if( abs( temp ) < abs(min - target) ) min = sum;
}
}

return min;
}
};

public class Solution {

// null
public int threeSumClosest(int[] num, int target) {
// Start typing your Java solution below
// DO NOT write main() function

// initial check.
//if( num==null || num.length<3) return 0;//throw new Exception("wrong input. must be at least 3 integers.");
int sz = num.length;

Arrays.sort( num );

int min = Integer.MAX_VALUE;  // pay attention. min's initial value should be MAX_VALUE.
int gsum = 0;    // since two possible sum (+-target) exist, we need gsum to record which sum.
for(int i=0; i<sz-2; i++) {
int j=i+1, k=sz-1;

while(j<k) {
int sum = num[i] + num[j] + num[k];
int x = Math.abs(sum-target);
if(x<min) {
min = x;
gsum = sum;
}

if(sum==target) {
++j;
--k;
return sum;
} else if( sum<target) {
++j;
} else {
--k;
}
}
}

return gsum;
}
}