leetcode 16. 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).

class Solution {struct Info{vector<pair<int, int>>pre;//pair<int,int>里的first代表sinnums里的index,second代表个数  int remainsize;//剩余的数字个数  int remainvalue;//剩余的值  };bool choose_one(int&k, vector<Info>&candi, map<int, int>&count, vector<int>sinnums){vector<Info>newcandi;if (k == 1){k--;for (int i = 0; i < candi.size(); i++)//考虑0  if (candi[i].remainvalue%candi[i].remainsize == 0){int nxtnum = candi[i].remainvalue / candi[i].remainsize;if (nxtnum>sinnums[candi[i].pre.back().first] && count.find(nxtnum) != count.end()&& count[nxtnum] >= candi[i].remainsize)return true;}}else{for (int i = 0; i < candi.size(); i++){for (int j = 1; j <= (candi[i].remainsize - k + 1); j++){for (int h = candi[i].pre.back().first + 1; (h < sinnums.size() - k + 1); h++){//粗略做判断，减小候选集大小  if (candi[i].remainvalue <= sinnums[h] * j + sinnums.back()* (candi[i].remainsize - j) && candi[i].remainvalue >= sinnums[h] * j + sinnums[h + 1] * (candi[i].remainsize - j) && j <= count[sinnums[h]]){Info info;info.pre = candi[i].pre;info.pre.push_back(pair<int, int>(h, j));info.remainsize = candi[i].remainsize - j;info.remainvalue = candi[i].remainvalue - j*sinnums[h];newcandi.push_back(info);}}}}k--;candi = newcandi;}return false;}bool threeSum(vector<int>& nums, int target){if (nums.empty())return false;map<int, int>count;for (int i = 0; i < nums.size(); i++)count[nums[i]]++;map<int, int>::iterator it = count.end(); it--;if (target >= 0 && count.begin()->first> target|| target < 0 && it->first < target)return false;vector<int>sinnums;for (it = count.begin(); it != count.end(); it++)sinnums.push_back(it->first);int neednum = 3;//答案里只有一种数字  if (target%neednum == 0 && count.find(target / neednum) != count.end() && count[target / neednum] >= neednum)return true;for (int i = 0; i < sinnums.size(); i++){//答案里有k种数字,2<=k<=neednum  for (int k = 2; k <= neednum; k++){if (sinnums.size() - i >= k){int jjmax = neednum - k + 1 < count[sinnums[i]] ? neednum - k + 1 : count[sinnums[i]];for (int j = 1; j <= jjmax; j++){Info inf;inf.pre.push_back(pair<int, int>(i, j));inf.remainsize = neednum - j;inf.remainvalue = target - j*sinnums[i];vector<Info>candi;candi.push_back(inf);int kk = k - 1;bool f = false;while (kk > 0 && !f)f = choose_one(kk, candi, count, sinnums);if (f)return true;}}}}return false;}public:int threeSumClosest(vector<int>& nums, int target){int pp = 0;bool f = threeSum(nums, target);if (f)return target;while (true){f = threeSum(nums, target + pp);if (f)return target + pp;f = threeSum(nums, target - pp);if (f)return target - pp;pp++;}};};

acceped

还是借上题的代码。。。

你可以说我是刷博客吧