LeetCode_3Sum Closest
一.题目
3Sum Closest
Total Accepted: 32191 Total Submissions: 119262My SubmissionsGiven 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).
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二.解题技巧
这道题和另外一道题3Sum类似,主要流程也是一样的。也是先进行排序,然后依照顺序选择数组A中的第i个元素作为三个数的最小值来寻找三个数的和。只是的地方在于这里是寻找最靠近给定值,寻找最靠近的值就无全部反复的事情了。所以能够不考虑夹逼的过程中的越过同样元素的过程,尽管越过同样的元素速度会快一些,可是代码长度也会加长。
这道题难的地方可能在于刚開始这样的差的阈值的过程。假设把阈值设置得太小了,会出现错误,因此,应该尽可能地将阈值设置得大一点。因为数组是已经排序的,因此,数组中三个数的和的范围在[3*A[0], 3*A[n-1]],因此,阈值能够依据以下三种情况进行设置:
这样就能够依据阈值与眼下得到的三个数的和与target的差来推断是否是最接近target的情况了,依据不同的情况,选择缩放的方向。1.if target >= 3*A[n-1],阈值设置为H = target - 3 * A[0];2.if 3*A[0] <= target<3*A[n-1],阈值设置为H = 3 * A[n-1] - 3*A[0];3.if target < 3 * A[0],阈值设置为H = 3 * A[n-1] - target。
三.实现代码
class Solution { public: int threeSumClosest(vector<int> &num, int target) { int Size = num.size(); sort(num.begin(), num.end()); int MaxSum = 3 * num[Size - 1]; int MinSum = 3 * num[0]; int ThreadHold = 0; if (target <= MinSum) { ThreadHold = MaxSum - target; } if (MaxSum < target) { ThreadHold = target - MinSum; } if ((MinSum < target) && (target <= MaxSum)) { ThreadHold = MaxSum - MinSum; } int Result = 0; for (int Index_outter = 0; Index_outter < (Size - 2); Index_outter++) { int First = num[Index_outter]; int Second = num[Index_outter + 1]; if ((Index_outter != 0) && (First == num[Index_outter - 1])) { continue; } int Start = Index_outter + 1; int End = Size - 1; while (Start < End) { Second = num[Start]; int Third = num[End]; int Sum = First + Second + Third; if (Sum == target) { return Sum; } if (Sum < target) { Start++; if (ThreadHold >= (target - Sum)) { Result = Sum; ThreadHold = target - Sum; } } if (Sum > target) { End--; if (ThreadHold >= (Sum - target)) { Result = Sum; ThreadHold = Sum - target; } } } } return Result; } };
四.体会
这道题最难的地方在于阈值的选择上面。事实上能够设置为整数的最大值的,可是,我一開始并不知道怎样计算整数的最大值,因此。仅仅能依据排好序的数组的三个数的和的范围与target的关系来设定阈值了,详细的阈值设置情况能够画个数轴出来分析,画出数轴之后,一切就明显了。
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