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).

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

解题思路：最直观的想法，三次循环，时间复杂度o(n^3),显然不能接受。

　　　　想到剪枝，在2次循环时，有很多情况是不必要走的，即如果得知num[i] + num[j] + num[k] > target的情况下，表明num[i] + num[m] + num[k] > target，其中m > j，同理还有小于的情况，于是复杂度可以降为o(n^2)