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  • [leedcode 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.
    • Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
    • 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 {
        List<Integer> seq;
        List<List<Integer>> res;
        public List<List<Integer>> combinationSum2(int[] candidates, int target) {
            //和之前的Combination Sum的不同是,这道题每个数字只可以使用一次。因此递归里传的值是i+1(2)
            //因为要进行去重,因此在结果中add时,需要判断结果中是否已经有值了(1)。level代表每次查找时的起始下标
            Arrays.sort(candidates);
            seq=new ArrayList<Integer>();
            res=new ArrayList<List<Integer>>();
            find(candidates,target,0,0);
    
            return res;
        }
        public void find(int[] candidates,int target,int sum,int level){
            if(sum==target){//1
                if(!res.contains(seq))
                res.add(new ArrayList<Integer>(seq));
                return;
            }
            if(sum>target){
                return;
            }
            for(int i=level;i<candidates.length;i++){
                seq.add(candidates[i]);
                sum+=candidates[i];
                find(candidates,target,sum,i+1);//2
                seq.remove(seq.size()-1);
                sum-=candidates[i];
            }
            
            
            
        }
    }
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  • 原文地址:https://www.cnblogs.com/qiaomu/p/4635987.html
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