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).

Solution：JAVA

public class Solution {
public int threeSumClosest(int[] num, int target) {
int min = Integer.MAX_VALUE;
int result = 0;

Arrays.sort(num);

for (int i = 0; i < num.length; i++) {
int j = i + 1;
int k = num.length - 1;
while (j < k) {
int sum = num[i] + num[j] + num[k];
int diff = Math.abs(sum - target);
if (diff < min) {
min = diff;
result = sum;
}
if (sum <= target) {
j++;
} else {
k--;
}
}
}

return result;
}
}

C++

class Solution {
public:
int threeSumClosest(vector<int> &num, int target) {
// Note: The Solution object is instantiated only once and is reused by each test case.
if(num.empty()) return 0;

sort(num.begin(), num.end());
int min= INT_MAX;
int record;

for(int i=0; i<num.size(); i++)
{
int tmp = target - num[i];
int start = i+1, end = num.size()-1;
while(start<end)
{
int sum = num[start]+num[end]+num[i];
if(sum==target)
{
min = 0;
record = sum;
break;
}
else if(sum > target)
{
if(abs(target-sum)<min)
{
min = abs(target-sum);
record = sum;
}
end--;
}
else if(sum < target)
{
if(abs(target-sum)<min)
{
min = abs(target-sum);
record = sum;
}
start++;
}
while(i<num.size()-1&&num[i]==num[i+1]) i++;
}
}
return record；
}
};

对数组求最大值，联想预处理排序