3Sum Closest - 每日算法

每日算法——leetcode系列

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.

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).

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;    }};