3Sum Closest(离目标值最近的三数之和)

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    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:(1)对数组进行排序。(2)遍历数组中的每个元素。设置两个指针,start指向该元素后一个位置,last指向数组的末尾。(3)求出这三个数之和与目标值得差值,如果当前差值小于上一次的差值,则把当前差值更新为最小的差值。同时更新最后结果为三数之和。最后返回结果即可。
int threeSumClosest(vector<int>& nums, int target){//最近距离int closestNum = INT_MAX;//三个数之和int res = 0;//当数组长度小于三时,返回最大整数if (nums.size() < 3)return closestNum;//对数组进行排序sort(nums.begin(), nums.end());//遍历数组for (int i = 0; i < nums.size() - 2; ++i){int start = i + 1;int last = nums.size() - 1;while (start < last){//当前三个数之和与目标值得差值int dis = nums[i] + nums[start] + nums[last] - target;//如果小于上一次的距离值,则更新最小距离和三个数之和if (abs(dis) < closestNum){closestNum = abs(dis);res = nums[i] + nums[start] + nums[last];}if (dis < 0){start++;}else if (dis > 0){last--;}else{return res;}}}return res;}
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