[LeetCode] (medium) 16. 3Sum Closest

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本文详细解析了LeetCode上的3Sum Closest问题，通过双指针法寻找数组中三个数的组合，使它们的和最接近给定的目标值。对比两种解决方案，强调了细节优化的重要性，并提供了正确的代码实现。

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

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.

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).

题目本身很简单，使用和3sum同样的方法就可以，问题的重点同样在与细节的优化

但是注意，在看了最快的提交代码之后发现其中有一处错误，他在最外层循环中直接使用了 if (nums[k] - target > dist) break;来剪枝，乍看之下好像没什么问题（nums[k]已经大于了当前解的上界，而[i][j]只会比[k]更大），但是细想之后就会发现当tar+dis<0时并不能保证后续不会出现更优解（反例：[-5,-4,-3,-2,0], -6，当k=1时）。

我的代码：

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        int re_dif = INT_MAX;

        int tem;
        
        int len = nums.size();
        
        int cur;
        
        sort(nums.begin(), nums.end());
        
        for(int i = 0; i < len; ++i){
            cur = nums[i]-target;
            // cout << i << ' ' << cur << endl;
            int lo = i+1;
            int hi = len-1;
            
            while(lo < hi){
                tem = cur+nums[lo]+nums[hi];
                // cout << i << ' ' << lo << ' ' << hi << ' ' << tem << endl;
                if(abs(tem) < abs(re_dif)) re_dif = tem;
                if(tem > 0) --hi;
                else if(tem < 0) ++lo;
                else{
                    return target;
                }
            }
            
        }
        
        return target+re_dif;
    }
};

问题代码：

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        int dist = INT_MAX, res;
        sort(nums.begin(), nums.end());
        if (nums.size() < 3) return 0;
        for (int k = 0; k < nums.size()-2; ++k) {
            if (nums[k] - target > dist) break;    //就是这一句判断有问题
            if (k > 0 && nums[k] == nums[k - 1]) continue;
            int i = k + 1, j = nums.size() - 1, sum;
            while (i < j) {
                sum = nums[k] + nums[i] + nums[j];
                if (sum == target)
                    return sum;
                else {
                    if (abs(sum - target) < dist) {
                        dist = abs(sum - target);
                        res = sum;
                    }
                    if (sum < target) ++i;
                    else --j;
                }
            }
        }
        return res;
    }
};