BackTracking

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);
                }
                
            }
        }
    }
}
原文地址:https://www.cnblogs.com/joannacode/p/5852568.html