标签：ons uniq blog back ack sha contain stop int
题目：
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:
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] ]
题解：
Solution 1 (TLE)
class Solution { public: void dfs(vector<vector<int>>& vv, vector<int>& v, vector<int> candidates, int target, int sum, vector<int>& visited) { if(sum == target) { vector<int>* tmp = new vector<int>; *tmp = v; sort((*tmp).begin(), (*tmp).end()); if(find(vv.begin(), vv.end(), *tmp) == vv.end()) vv.push_back(*tmp); delete tmp; return; } if(sum > target) return; for(int i=0; i<candidates.size(); ++i) { if(visited[i] != 0) continue; v.push_back(candidates[i]); visited[i] = 1; dfs(vv, v, candidates, target, sum+candidates[i], visited); v.pop_back(); visited[i] = 0; } } vector<vector<int>> combinationSum2(vector<int>& candidates, int target) { vector<vector<int>> vv; vector<int> v; vector<int> visited(candidates.size(),0); dfs(vv, v, candidates, target, 0, visited); return vv; } };
Solution 1 中即使i从level开始遍历，也无法accepted，加入stop后才勉强通过，即Solution 2
Solution 2 (almost TLE but not 579ms)
class Solution { public: void dfs(vector<vector<int>>& vv, vector<int>& v, vector<int> candidates, int target, int sum, vector<int>& visited, int stop,int level) { if(sum == target) { vector<int>* tmp = new vector<int>; *tmp = v; sort((*tmp).begin(), (*tmp).end()); if(find(vv.begin(), vv.end(), *tmp) == vv.end()) vv.push_back(*tmp); delete tmp; return; } if(sum > target) {stop = 1;return;} for(int i=level; i<candidates.size(); ++i) { if(visited[i] != 0) continue; v.push_back(candidates[i]); visited[i] = 1; dfs(vv, v, candidates, target, sum+candidates[i], visited, stop, level+1); if(stop == 1) {stop = 0;break;} v.pop_back(); visited[i] = 0; } } vector<vector<int>> combinationSum2(vector<int>& candidates, int target) { vector<vector<int>> vv; vector<int> v; vector<int> visited(candidates.size(),0); sort(candidates.begin(), candidates.end()); dfs(vv, v, candidates, target, 0, visited, 0, 0); return vv; } };
Solution 3 ()
class Solution { public: void dfs(vector<vector<int>>& vv, vector<int>& v, vector<int> candidates, int target, vector<int>& visited, int stop,int level) { if(target == 0) { vector<int>* tmp = new vector<int>; *tmp = v; sort((*tmp).begin(), (*tmp).end()); if(find(vv.begin(), vv.end(), *tmp) == vv.end()) vv.push_back(*tmp); delete tmp; return; } if(target<0) {stop = 1;return;} for(int i=level; i<candidates.size(); ++i) { if(visited[i] != 0) continue; v.push_back(candidates[i]); visited[i] = 1; dfs(vv, v, candidates, target-candidates[i], visited, stop,level+1); if(stop == 1) {stop = 0;break;} v.pop_back(); visited[i] = 0; } } vector<vector<int>> combinationSum2(vector<int>& candidates, int target) { vector<vector<int>> vv; vector<int> v; vector<int> visited(candidates.size(),0); sort(candidates.begin(), candidates.end()); dfs(vv, v, candidates, target, visited, 0,0); return vv; } };
Solution 4
【LeetCode】040. Combination Sum II
标签：ons uniq blog back ack sha contain stop int
原文地址：http://www.cnblogs.com/Atanisi/p/6789004.html
