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    BackTracking (DFS)

    39. Combination Sum

    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.
    • 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]
    ]
    public class Solution {
        public List<List<Integer>> combinationSum(int[] candidates, int target) {
            List<List<Integer>> res = new ArrayList<>();
            List<Integer> member = new ArrayList<Integer>();
            helper(res, member, candidates, target, 0);
            return res;
        }
        public void helper(List<List<Integer>> res, List<Integer> member, int[] candidates, int target, int start){
           if(target < 0)
                return;
           else if(target == 0){
                res.add(new ArrayList<Integer>(member)); //member is address
                return;
            }
            else{
                for(int i = start; i < candidates.length; i++){
                    member.add(candidates[i]);
                    helper(res, member, candidates, target - candidates[i], i );
                    member.remove(member.size() - 1); //make member empty!
                }
            }
        }
    }

    40. 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.
    • 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]
    ]
    public class Solution {
        public List<List<Integer>> combinationSum2(int[] candidates, int target) {
            List<List<Integer>> res = new ArrayList<>();
            List<Integer> member = new ArrayList<>();
            Arrays.sort(candidates);
            boolean visit[] = new boolean[candidates.length];
            helper(res, member, visit, candidates, target, 0);
            return res;
        }
        public void helper(List<List<Integer>> res, List<Integer> member ,boolean[] visit, int[] candidates , int target ,int deep){
            if(target < 0)
                return ;
            else if(target == 0){
                res.add(new ArrayList<Integer>(member));
                return;
            }
            else{
                for(int i = deep; i < candidates.length; i++){
                    if(!visit[i]){
                        if (i > 0 && candidates[i] == candidates[i-1] && visit[i-1]==false) continue;
                        member.add(candidates[i]);
                        visit[i] = true;
                        helper(res, member, visit, candidates, target - candidates[i], i);
                        visit[i] = false;
                        member.remove(member.size() - 1);
                    }
                    
                }
            }
        }
    }
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  • 原文地址:https://www.cnblogs.com/joannacode/p/5852568.html
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