LeetCode 16 3Sum Closest（最接近的3个数的和）

翻译

给定一个有n个整数的数组S，找出S中3个数，使其和等于一个给定的数，target。

返回这3个数的和，你可以假定每个输入都有且只有一个结果。

例如，给定S = {-1 2 1 -4}，和target = 1。

那么最接近target的和是2。(-1 + 2 + 1 = 2)。

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

也许我已经开始体会到上一题中别人写的方法的思想了。

在这个题目中，我们要做以下几件事：

• 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;
    }
};