leetcode 40. Combination Sum II

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

Note:

• All numbers (including target) will be positive integers.
• Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
• The solution set must not contain duplicate combinations.

For example, given candidate set 10,1,2,7,6,1,5 and target 8, 
A solution set is: 
[1, 7] 
[1, 2, 5] 
[2, 6] 
[1, 1, 6]

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    };vector<vector<int>>re;void choose_one(int&k, vector<Info>&candi, map<int, int>&count, vector<int>sinnums){vector<Info>newcandi;if (k == 1){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){vector<int>aa;for (int j = 0; j < candi[i].pre.size(); j++)for (int h = 0; h < candi[i].pre[j].second; h++)aa.push_back(sinnums[candi[i].pre[j].first]);for (int j = 0; j < candi[i].remainsize; j++)aa.push_back(nxtnum);re.push_back(aa);}}k--;return;}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;}}public:vector<vector<int>> combinationSum2(vector<int>& candidates, int target){if (candidates.empty())return re;map<int, int>count;for (int i = 0; i < candidates.size(); i++)count[candidates[i]]++;map<int, int>::iterator it = count.end(); it--;if (target >= 0 && count.begin()->first> target|| target < 0 && it->first < target)return re;vector<int>sinnums;for (it = count.begin(); it != count.end(); it++)sinnums.push_back(it->first);int maxneednum = target/sinnums.front();int neednum;//答案里只有一种数字 for (neednum = 1; neednum <= maxneednum;neednum++)if (target%neednum == 0 && count.find(target / neednum) != count.end() && count[target / neednum] >= neednum){vector<int>aa(neednum, target / neednum);re.push_back(aa);}for (int i = 0; i < sinnums.size(); i++){//答案里有k种数字,2<=k<=neednum int need = target / sinnums[i];if (need >= 2){for (neednum = 2; neednum <= need; 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;while (kk > 0)choose_one(kk, candi, count, sinnums);}}}}}return re;}};

accepted