Leetcode 40. Combination Sum II

題目

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:
All numbers (including target) will be positive integers.
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

從第一個往前遞歸計算

測試用例

[] 3
[2] 2
[1, 1, 1, 2, 3, 4, 5] 8
[2] 0

代碼

package leetcodeArray;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.List;

public class Leetcode40CombinationSum2 {
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
    Arrays.sort(candidates);
    List<List<Integer>> result = new ArrayList<List<Integer>>();
    List<Integer> cur = new LinkedList<Integer>();;
    combination(candidates, target, result, cur, -1);
    return result;
}

private void combination(int[] candidate, int target, List<List<Integer>> result, List<Integer> cur, int start){
    if(target == 0){
        result.add(new LinkedList<Integer>(cur));
        return;
    }

    for(int i = start + 1; i < candidate.length && candidate[i] <= target;i++){
        cur.add(candidate[i]);
        combination(candidate, target - candidate[i], result, cur, i);
        cur.remove(cur.size() - 1);
        while(i < candidate.length - 1 && candidate[i] == candidate[i+1]) i++;
    }

}
}

結果

他山之玉

public List<List<Integer>> combinationSum2(int[] cand, int target) {
    Arrays.sort(cand);
    List<List<Integer>> res = new ArrayList<List<Integer>>();
    List<Integer> path = new ArrayList<Integer>();
    dfs_com(cand, 0, target, path, res);
    return res;
}
void dfs_com(int[] cand, int cur, int target, List<Integer> path, List<List<Integer>> res) {
    if (target == 0) {
        res.add(new ArrayList(path));
        return ;
    }
    if (target < 0) return;
    for (int i = cur; i < cand.length; i++){
        if (i > cur && cand[i] == cand[i-1]) continue;
        path.add(path.size(), cand[i]);
        dfs_com(cand, i+1, target - cand[i], path, res);
        path.remove(path.size()-1);
    }
}

class Solution {
public:
    vector<vector<int> > combinationSum2(vector<int> &num, int target) 
    {
        vector<vector<int>> res;
        sort(num.begin(),num.end());
        vector<int> local;
        findCombination(res, 0, target, local, num);
        return res;
    }
    void findCombination(vector<vector<int>>& res, const int order, const int target, vector<int>& local, const vector<int>& num)
    {
        if(target==0)
        {
            res.push_back(local);
            return;
        }
        else
        {
            for(int i = order;i<num.size();i++) // iterative component
            {
                if(num[i]>target) return;
                if(i&&num[i]==num[i-1]&&i>order) continue; // check duplicate combination
                local.push_back(num[i]),
                findCombination(res,i+1,target-num[i],local,num); // recursive componenet
                local.pop_back();
            }
        }
    }
};

def combinationSum2(self, candidates, target):
    candidates.sort()
    table = [None] + [set() for i in range(target)]
    for i in candidates:
        if i > target:
            break
        for j in range(target - i, 0, -1):
            table[i + j] |= {elt + (i,) for elt in table[j]}
        table[i].add((i,))
    return map(list, table[target])