leetcode | 3Sum Closest

3Sum Closest : https://leetcode.com/problems/3sum-closest/

问题描述

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).

解析

和 上篇 3Sum 思想相同，利用夹逼思想，枚举一个数nums[i]，从j=i+1, k = nums.size()-1;夹逼，sum小于target：j++，sum大于target，k–
此问题不需考虑重复，因为只需要一个和的结果。

class Solution {public:    int threeSumClosest(vector<int>& nums, int target) {        int result = nums[0]+nums[1]+nums[2];//初始化        sort(nums.begin(), nums.end());        for (int i = 0; i < nums.size()-2; i++) {            int j = i+1;            int k = nums.size()-1;            int sum;            while (j < k) {                sum = nums[i]+nums[j]+nums[k];                if (abs(sum-target) < abs(result-target))                    result = sum;                if (sum < target)                    j++;  //不需检测nums[j]==nums[j-1],可以重复，不需保存满足条件的元素                else if (sum > target)                    k--;                else                    return target;            }        //注：不可将第12、13行放在此处，因为此处的终止条件是j == k        }        return result;      }};