9.8 甩硬币方式,leetcode combine sum I II

本文提供LeetCode上两道经典题目“组合求和”与“组合求和II”的解决方案,采用递归方法实现,重点介绍如何避免重复组合及确保结果集不包含重复项的技术细节。

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leetcode combine sum I

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.
  • Elements in a combination (a1a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
  • 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 ArrayList<ArrayList<Integer>> combinationSum(int[] candidates, int target) {
        if(candidates==null||candidates.length==0) return null;
        Arrays.sort(candidates);
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        helper(res, candidates, target, new ArrayList<Integer>(), 0);
        return res;
    }
    
    private void helper(ArrayList<ArrayList<Integer>> res, int[] cand, int target, ArrayList<Integer> item, int start){
        if(target<0) return;
        if(target==0){
            res.add(new ArrayList<Integer> (item));  //think why must add new here ?? 除了基本8种data types,http://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html, 其他java里的类型都是指针,此处传入的item为指针, 在之后的引用中变化了,所以如果不new 一个新的copy,会在这里也被错误的变化了
            
        }
        
        for(int i=start;i<cand.length;i++){
            if(i>start && cand[i]==cand[i-1]) continue; // which means this candidate has been examined;
            item.add(cand[i]);
            helper(res, cand, target-cand[i], item, i);
            item.remove(item.size()-1);
        }
        
    }

}

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 (a1a2, … , 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 {


    public ArrayList<ArrayList<Integer>> combinationSum2(int[]  num, int target) {
        
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        if(num.length==0||num==null) return res;
        Arrays.sort(num);
        helper(0, num, target, res, new ArrayList<Integer>());
        return res;
    }
    
    private void helper(int start, int[] num, int target, ArrayList<ArrayList<Integer>> res,ArrayList<Integer> item){
   
         if(target<0) {
            return;
        }
            
        if(target==0){
            res.add(new ArrayList<Integer>(item));  
            return;
        }
 
        
        for(int i=start;i<num.length;i++){
            if(i>start&&num[i]==num[i-1]) continue;
            item.add(num[i]);
            helper(
i+1,num,target-num[i], res,item);
            item.remove(item.size()-1);
        }
    }
}





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