16. 3Sum Closest

最新推荐文章于 2019-05-28 15:42:21 发布

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最新推荐文章于 2019-05-28 15:42:21 发布

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原文链接：http://www.cnblogs.com/forPrometheus-jun/p/10656867.html

本文介绍了一个算法问题，即寻找数组中三个整数的组合，使其和最接近给定的目标值。通过详细解释解决方案，包括排序数组和使用双指针技巧，来逐步逼近最优解。

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description:

Given an array nums of n integers and an integer target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

Note:

Example:

Given array nums = [-1, 2, 1, -4], and target = 1.

The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

my answer:

感恩

my answer

思路和上一道题3sum是一样的，就是多加了两个变量closest&diff,closest记录答案，diff记录最小差值

大佬的answer：

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        int closest = nums[0] + nums[1] + nums[2];
        int diff = abs(target - closest);
        sort(nums.begin(),nums.end());
        for (int i = 0; i < nums.size() - 2; ++i){
            int left = i + 1;
            int right = nums.size() - 1;
            while(left < right){
                int sum = nums[i] + nums[left] + nums[right];
                int newdiff = abs(target - sum);
                if(diff > newdiff){
                    diff = newdiff;
                    closest = sum;
                }
                if(sum < target) left++;
                else right--;
            }
            
        }
        return closest;
    }
};