标签：one abs -- ret number interview nbsp color neu
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).
此题和之前两个three sum的区别在于，此题是求最接近target的值的三个数的和，和之前three sum一样，利用排序，for循环，然后双指针来解决此题，不同之处在于创建一个sum值，每次更新最小三个数组的和。代码如下：
public class Solution {
public int threeSumClosest(int[] nums, int target) {
Arrays.sort(nums);
int len = nums.length;
int sum = nums[0]+nums[1]+nums[2];
for(int i=0;i<len-2;i++){
int low = i+1,high = len-1;
while(low<high){
int cur = nums[i]+nums[low]+nums[high];
if(cur<=target){
if(Math.abs(sum-target)>Math.abs(cur-target)){
sum = cur;
}
low++;
}else{
if(Math.abs(sum-target)>Math.abs(cur-target)){
sum = cur;
}
high--;
}
}
}
return sum;
}
}
标签：one abs -- ret number interview nbsp color neu
原文地址：http://www.cnblogs.com/codeskiller/p/6354103.html
