Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.
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
A solution set is:
2,3,6,7
and target 7
,A solution set is:
[7]
[2, 2, 3]
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
A solution set is:
10,1,2,7,6,1,5
and target 8
,A solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]
思路:Combination Sum I
和3 sum那题很像,如果选定一个candidates[i],则需要继续寻找和为target-candidate[i]的combination。由于candidates和target都为正数,当和超过或等于target时,查找中止。与3sum的思路一样,对candidate排序,并且每层递归扫描的时候需要做到去重复解:
1. 不回头扫,在扫描candidates[i]时,对candidate[i: n-1]递归查找target-candidates[i]。
2. 每层扫描的时候跳过重复的candidates。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 | class Solution { public: vector<vector<int> > combinationSum(vector<int> &candidates, int target) { vector<vector<int> > allSol; vector<int> sol; if(candidates.empty()) return allSol; sort(candidates.begin(),candidates.end()); findCombSum(candidates, 0, target, sol, allSol); return allSol; } void findCombSum(vector<int> &candidates, int start, int target, vector<int> &sol, vector<vector<int>> &allSol) { if(target==0) { allSol.push_back(sol); return; } for(int i=start; i<candidates.size(); i++) { if(i>start && candidates[i]==candidates[i-1]) continue; if(candidates[i]<=target) { sol.push_back(candidates[i]); findCombSum(candidates, i, target-candidates[i], sol, allSol); sol.pop_back(); } } } }; |
思路:Combination Sum II
和I的唯一区别是每个candidate只能用一次。只需要在调用下一层递归时,查找范围的起始数字不是当前数字而是下一个数字即可。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 | class Solution { public: vector<vector<int> > combinationSum2(vector<int> &num, int target) { vector<vector<int> > allSol; if(num.empty()) return allSol; sort(num.begin(),num.end()); vector<int> sol; findCombSum2(num, 0, target, sol, allSol); return allSol; } void findCombSum2(vector<int> &num, int start, int target, vector<int> &sol, vector<vector<int> > &allSol) { if(target==0) { allSol.push_back(sol); return; } for(int i=start; i<num.size(); i++) { if(i>start && num[i]==num[i-1]) continue; if(num[i]<=target) { sol.push_back(num[i]); findCombSum2(num, i+1, target-num[i], sol, allSol); sol.pop_back(); } } } }; |
要剪枝叶啊,
ReplyDeleteif(num[i]<=target) ...
else break;