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).
同上题，没上一题难，比葫芦画瓢即可
JAVA实现：
static public int threeSumClosest(int[] nums, int target) {
int result = nums[0] + nums[1] + nums[2];
Arrays.sort(nums);
for (int i = 0; i < nums.length - 2; i++) {
int j = i + 1, k = nums.length - 1;
while (j < k) {
if (Math.abs(nums[i] + nums[j] + nums[k] - target) < Math.abs(result - target))
result = nums[i] + nums[j] + nums[k];
if (nums[i] + nums[j] + nums[k] < target)
j++;
else if (nums[i] + nums[j] + nums[k] > target)
k--;
else
return target;
//不加while循环亦可通过测试，但是时间会偏长一些
while (i < k && nums[i] == nums[i + 1])
i++;
}
}
return result;
}
C++:
class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
int result = nums[] + nums[] + nums[];
sort(nums.begin(),nums.end());
for (int i = ; i < nums.size() - ; i++) {
int j = i + , k = nums.size() - ;
while (j < k) {
if (abs(nums[i] + nums[j] + nums[k] - target) < abs(result - target))
result = nums[i] + nums[j] + nums[k];
if (nums[i] + nums[j] + nums[k] < target)
j++;
else if (nums[i] + nums[j] + nums[k] > target)
k--;
else
return target;
while (i < k && nums[i] == nums[i + ])
i++;
}
}
return result;
}
};