标签:style color io os java ar for art div
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).
public class Solution {
public void swap(int[] array, int i, int j)
{
int t = array[i];
array[i] = array[j];
array[j] = t;
}
public void quickSort(int[] array, int start, int end)
{
if (start >= end)
{
return;
}
int middle = (start + end) / 2;
swap(array, start, middle);
int i = start + 1;
while (i <= end)
{
if (array[i] >= array[start])
{
break;
}
i++ ;
}
int j = i + 1;
for (; j <= end;)
{
while (j <= end)
{
if (array[j] < array[start])
{
break;
}
j++ ;
}
if (j <= end)
{
swap(array, i, j);
i++ ;
j++ ;
}
}
i-- ;
if (array[i] < array[start])
{
swap(array, start, i);
}
quickSort(array, start, i - 1);
quickSort(array, i + 1, end);
}
public int threeSumClosest(int[] num, int target) {
int result=num[0]+num[1]+num[2];
int diff=Math.abs(result-target);
int sum;
int newDiff;
int i;
int j;
int k;
quickSort(num,0,num.length-1);
for(i=0;i<num.length-2;i++){
j=i+1;
k=num.length-1;
while(j<k)
{
sum=num[i]+num[j]+num[k];
newDiff=Math.abs(sum-target);
if(newDiff<diff){
diff=newDiff;
result=sum;
}
if(sum<target)
{
j++;
}
else{
k--;
}
}
}
return result;
}
}标签:style color io os java ar for art div
原文地址:http://blog.csdn.net/jiewuyou/article/details/39288681