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# 3 Sum Closest（最接近的 3 个数的和）

## 原文

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).

## 思考

• 用 sort 对输入的数组进行排序
• 求出长度 len，current 之所以要小于 len−2，是因为后面需要留两个位置给 front 和 back
• 始终保证 front 小于 back
• 计算索引为 current、front、back 的数的和，分别有比 target 更小、更大、相等三种情况
• 更小：如果距离小于 close，那么close 便等于 target−sum，而结果就是 sum。更大的情况同理
• 如果相等，那么要记得将 0 赋值给 close，result 就直接等于 target 了
• 随后为了避免计算重复的数字，用三个 do/while 循环递增或递减它们

## 代码

class Solution
{
public:
int threeSumClosest(vector<int>& nums, int target) {
sort(nums.begin(), nums.end());
int len = nums.size();
int result = INT_MAX, close = INT_MAX;
for (int current = 0; current < len - 2; current++) {
int front = current + 1, back = len - 1;
while (front < back) {
int sum = nums[current] + nums[front] + nums[back];
if (sum < target) {
if (target - sum < close) {
close = target - sum;
result = sum;
}
front++;
}
else if (sum > target) {
if (sum - target < close) {
close = sum - target;
result = sum;
}
back--;
}
else {
close = 0;
result = target;
do {
front++;
} while (front < back&&nums[front - 1] == nums[front]);
do {
back--;
} while (front < back&&nums[back + 1] == nums[back]);
}
}
while (current < len - 2 && nums[current + 1] == nums[current]) {
current++;
}
}
return result;
}
};