Monday, November 17, 2014

[LeetCode] Combination Sum I, II

Combination Sum I

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 (a1a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
  • The solution set must not contain duplicate combinations.
For example, given candidate set 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 (a1a2, … , 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] 

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。


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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。


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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();
            }
        }
    }
};

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