Leetcode 40. Combination Sum II

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:
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

从第一个往前递归计算

测试用例

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

代码

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++;    }}}

结果

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])