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  • 39. Combination Sum (Java)

    Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

    The same repeated number may be chosen from candidates unlimited number of times.

    Note:

    • All numbers (including target) will be positive integers.
    • The solution set must not contain duplicate combinations.

    Example 1:

    Input: candidates = [2,3,6,7], target = 7,
    A solution set is:
    [
      [7],
      [2,2,3]
    ]
    

    Example 2:

    Input: candidates = [2,3,5], target = 8,
    A solution set is:
    [
      [2,2,2,2],
      [2,3,3],
      [3,5]
    ]
    class Solution {
        public List<List<Integer>> combinationSum(int[] candidates, int target) {
            List<Integer> ans = new ArrayList<Integer>();
            Arrays.sort(candidates);
            backTrack(candidates, target, 0, ans, 0);
            return result;
        }
        
        public void backTrack(int[] candidates, int target, int start, List<Integer> ans, int sum){
            if(start >= candidates.length || sum + candidates[start] > target) 
                return; //not found
            else if(sum + candidates[start] == target ){ //found an answer
                List<Integer> new_ans = new ArrayList<Integer>(ans); //不能用List<Integer> new_ans = ans;这个只是创建了原List的一个引用
                new_ans.add(candidates[start]);
                result.add(new_ans);
            }
            else{
                // not choose current candidate
                backTrack(candidates,target,start+1,ans,sum);
                    
                //choose current candidate
                ans.add(candidates[start]);
                backTrack(candidates,target,start,ans,sum+candidates[start]);
                ans.remove(ans.size()-1); //List是按引用传递,为了不影响递归,需要复原
            }
        }
        
        private List<List<Integer>> result = new ArrayList<List<Integer>>();
    }

    注意List按引用(地址)传递,需要新new一个,或是复原,以不影响其他部分

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  • 原文地址:https://www.cnblogs.com/qionglouyuyu/p/10812822.html
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