题目
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).
解题之法
class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
int closest = nums[0] + nums[1] + nums[2];
int diff = abs(closest - target);
sort(nums.begin(), nums.end());
for (int i = 0; i < nums.size() - 2; ++i) {
int left = i + 1, right = nums.size() - 1;
while (left < right) {
int sum = nums[i] + nums[left] + nums[right];
int newDiff = abs(sum - target);
if (diff > newDiff) {
diff = newDiff;
closest = sum;
}
if (sum < target) ++left;
else --right;
}
}
return closest;
}
};
分析
这道题让我们求最接近给定值的三数之和,是在之前那道 3Sum 三数之和的基础上又增加了些许难度,那么这道题让我们返回这个最接近于给定值的值,即我们要保证当前三数和跟给定值之间的差的绝对值最小,所以我们需要定义一个变量diff用来记录差的绝对值,然后我们还是要先将数组排个序,然后开始遍历数组,思路跟那道三数之和很相似,都是先确定一个数,然后用两个指针left和right来滑动寻找另外两个数,每确定两个数,我们求出此三数之和,然后算和给定值的差的绝对值存在newDiff中,然后和diff比较并更新diff和结果closest即可。