【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).

java code : 复杂度 O(n^2 logn) 最内层用二分查找

public class Solution {
public int threeSumClosest(int[] num, int target) {
// Note: The Solution object is instantiated only once and is reused by each test case.
if(num.length < 3)
return 0;
Arrays.sort(num);
int sum = num[0] + num[1] + num[2];
int dist = Math.abs((num[0] + num[1] + num[2] - target));
if(dist == 0)
return sum;
for(int i = 0; i < num.length - 2; i++)
{
for(int j = i + 1; j < num.length -1; j++)
{
int val = target - num[i] - num[j];
int index = binarysearch(num, val, j+1, num.length - 1);
if(num[index] == val)
return target;
else
{
int sum1 = num[i] + num[j] + num[index];
int dist1 = Math.abs((target - sum1));
if(dist1 < dist)
{
sum = sum1;
dist = dist1;
}
if(index < num.length -1)
{
int sum2 = num[i] + num[j] + num[index + 1];
int dist2 = Math.abs((target - sum2));
if(dist2 < dist)
{
sum = sum2;
dist = dist2;
}
}
}
}
}
return sum;
}
public int binarysearch(int[] array, int key, int st, int end)
{
int lhs = st, rhs = end;
if(st == end)
return st;
while(lhs <= rhs)
{
int mid = (lhs + rhs) >> 1;
if(array[mid] == key)
return mid;
else if(array[mid] < key)
lhs = mid + 1;
else rhs = mid - 1;
}
if(rhs < st)
return st;
return rhs;
}
}