标签:and item 之间 一个 max pre content code each
例如,给定的整数 S = {-1 2 1 -4}, 目标值 = 1.↵↵ 最接近目标值的和为 2. (-1 + 2 + 1 = 2).
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).
这道题让我们求最接近给定值的三数之和,是在之前那道 3Sum 的基础上又增加了些许难度,那么这道题让返回这个最接近于给定值的值,
即要保证当前三数和跟给定值之间的差的绝对值最小,所以需要定义一个变量 diff 用来记录差的绝对值,然后还是要先将数组排个序,
然后开始遍历数组,思路跟那道三数之和很相似,都是先确定一个数,然后用两个指针 left 和 right 来滑动寻找另外两个数,
每确定两个数,求出此三数之和,然后算和给定值的差的绝对值存在 newDiff 中,然后和 diff 比较并更新 diff 和结果 closest 即可,代码如下:
class Solution { public: int threeSumClosest(vector<int> &num, int target) { int res; int d = INT_MAX; sort(num.begin(),num.end()); int n = num.size(); for(int k=0;k<n;++k) { int t = target-num[k]; int i=k+1,j=n-1; while(i<j) { int m = t-num[i]-num[j]; if(abs(m)<abs(d)) d=m; if(m>0) i++; else j--; } } res=target-d; return res; } };
我们还可以稍稍进行一下优化,每次判断一下,当 nums[i]*3 > target 的时候,就可以直接比较 closest 和 nums[i] + nums[i+1] + nums[i+2] 的值,返回较小的那个,因为数组已经排过序了,后面的数字只会越来越大,就不必再往后比较了,参见代码如下:
class Solution { public: int threeSumClosest(vector<int>& nums, int target) { int closest = nums[0] + nums[1] + nums[2]; int diff = abs(closest - target); sort(nums.begin(), nums.end()); for (int i = 0; i < nums.size() - 2; ++i) { if (nums[i] * 3 > target) return min(closest, nums[i] + nums[i + 1] + nums[i + 2]); int left = i + 1, right = nums.size() - 1; while (left < right) { int sum = nums[i] + nums[left] + nums[right]; int newDiff = abs(sum - target); if (diff > newDiff) { diff = newDiff; closest = sum; } if (sum < target) ++left; else --right; } } return closest; } };
[LeetCode] 16. 3Sum Closest 最近三数之和
标签:and item 之间 一个 max pre content code each
原文地址:https://www.cnblogs.com/zl1991/p/12764321.html