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;}
};