和大神们学习每天一题(leetcode)-3Sum Closest

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).

本题和3Sum那道题类似，不同点在于需要计算与目标值的差距。

功能测试用例：{ -1, 2, 1, -4 },4

特殊测试用例：{},0;{0},0;{0,0},0

class Solution{public:void FastSort(vector<int> &vsNumber, int nBegin, int nEnd)//快速排序{if (nBegin >= nEnd)return;int nPos1 = nBegin;int sNMiddle = vsNumber[nEnd], sNChange;for (int nPos2 = nBegin; nPos2 < nEnd; nPos2++){if (vsNumber[nPos2] < sNMiddle){sNChange = vsNumber[nPos1];vsNumber[nPos1] = vsNumber[nPos2];vsNumber[nPos2] = sNChange;nPos1++;}}vsNumber[nEnd] = vsNumber[nPos1];vsNumber[nPos1] = sNMiddle;FastSort(vsNumber, nBegin, nPos1 - 1);FastSort(vsNumber, nPos1 + 1, nEnd);}int threeSumClosest(vector<int> &num, int target){if (num.size() < 3)return NULL;int nResult = NULL,nRemain,nBegin,nEnd,nClosest = INT32_MAX,nAbs;FastSort(num,0,num.size()-1);//对输入的向量中元素进行排序for (int nTemp = 0; nTemp < num.size() - 2; nTemp++)//遍历向量中的元素{nRemain = target - num[nTemp];//计算出减去当前值后的剩余值nBegin = nTemp + 1;nEnd = num.size() - 1;while (nBegin < nEnd){nAbs = abs(num[nBegin] + num[nEnd] - nRemain);//计算与目标值差值的绝对值if (nAbs < nClosest)//与最小绝对值进行比较{nClosest = nAbs;nResult = num[nBegin] + num[nEnd] + num[nTemp];}if (num[nBegin] + num[nEnd] < nRemain){nBegin++;}else if (num[nBegin] + num[nEnd] > nRemain){nEnd--;}else{return target;}}}return nResult;}};