3Sum Closest

每日算法——leetcode系列

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问题 3Sum Closest

Difficulty: Medium

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.

<pre>
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).

</pre>

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

翻译

三数和

难度系数：中等

给定一个有n个整数的数组S, 找出数组S中三个数的和最接近一个指定的数。 返回这三个数的和。 你可以假定每个输入都有一个结果。

思路

思路跟3Sum很像， 不同的就是不用去重复， 要记录三和最接近target的和值。

代码

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        int result = 0;
        int minDiff = INT32_MAX;
        // 排序
        sort(nums.begin(), nums.end());
        int n = static_cast<int>(nums.size());
        
        for (int i = 0; i < n - 2; ++i) {
          
            int j = i + 1;
            int k = n - 1;
            while (j < k) {
                int sum = nums[i] + nums[j] + nums[k];
                int diff = abs(sum - target);
                // 记录三和最接近target的和值
                if (minDiff > diff){
                    result = sum;
                    minDiff = diff;
                }
                else if (sum < target){
                    j++;
                }
                else if (sum > target){
                    k--;
                }
                else{
                    return result;
                }
            }
        }
        return result;
    }
};