原题如下：

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).

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思路如下：

找最接近的和，暴力枚举没种可能的情况，然后记录最接近的值，即和与目标值进行绝对值差之后的结果。

在枚举的时候一点技巧，比如1，2，3，4，5，6。这组数据，先对这组数据排序，然后num[1]+ （ num[2] + num[6]），然后

后续下表依次靠近，找完num[1]之后，在搜以num[2]为第一个元素的时候，则计算num[2]+ (num[3]+num[6]),不需要再管前面

的num[1],避免造成暴力的重复。

代码如下：

（java）

public class Solution {

/*public int threeSumClosest(int[] nums, int target) {

int rez = 0;

if(nums.length == 0) return 0;

else if(nums.length == 1) return nums[0];

else if(nums.length == 2) return (nums[0]+nums[1]);

else

{

int min = Integer.MAX_VALUE;

Arrays.sort(nums);

int k = nums.length-1;

for(int i = 0; i < nums.length; i++)

{

for(int j = i+1; j < k;)

{

int sum = nums[i]+nums[j]+nums[k];

int temp = Math.abs(sum-target);

if(temp == 0)

return sum;

else

{

if(temp < min)

{

min = temp;

rez = sum;

}

if(sum <= target)

j++;

if(sum > target)

k--;

}

}

}

return rez;

}

}

}*///不造为毛这段代码会wa。。然后我觉得跟标程明明差不多。。

public int threeSumClosest(int[] nums, int target) {

int min = Integer.MAX_VALUE;

int result = 0;

Arrays.sort(nums);

for (int i = 0; i < nums.length; i++) {

int j = i + 1;

int k = nums.length - 1;

while (j < k) {

int sum = nums[i] + nums[j] + nums[k];

int diff = Math.abs(sum - target);

if(diff == 0) return sum;

if (diff < min) {

min = diff;

result = sum;

}

if (sum <= target) {

j++;

} else {

k--;

}

}

}

return result;

}

}