LeetCode OJ 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.
  • 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 2,3,6,7 and target 7
A solution set is: 
[7] 
[2, 2, 3]

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【题目解析】

给定一个候选数字序列一个目标值,找到所有的和等于目标值的候选数字组合。同一个候选数字可以出现多次。

1. 所有的数字是正数;

2. 组合中的数字降序排列;

3. 结果中不能存在相同的组合;

【思路】

首先我们从一个简单的例子分析一下这样组合生成的过程:

[1,2,3] target = 7

我们从[1,2,3]生成所有的和等于7的数字组合,可以分为以1开头的有:

[1,1,1,1,1,1,1],[1,1,1,1,1,2],[1,1,1,1,3],[1,1,1,2,2],[1,1,2,3],[1,2,2,2],[1,3,3]

以2开头的:

[2,2,3]

上面这几个组合是满足所有条件的组合。生成过程可以这样描述:

1. 给定一个升序排序的数组;

2. 从数组头部向后遍历,如果遇到一个比目标值小的数n,我们找到目标值为target - n的所有组合,如果找到这样的组合,那么我们把n合并到每一个组合里;

3. 如果遇到一个值m = target 则新建一个List,添加m后返回;

4. 如果遇到一个值m > target 则终结遍历,因为之后的数字肯定比target还大;

很明显这是一个递归的过程,在这个过程中我们首先对候选数组进行了排序,这是为什么呢? 因为在递归的过程中,如果一个数字n小于target,那么在递归求解target - n时我们可以确定一个从候选数组重新开始的下标,这个下标就是当前数字的下标。

【java代码】

 1 public class Solution {
 2     public List<List<Integer>> combinationSum(int[] candidates, int target) {
 3         Arrays.sort(candidates);        //升序排序
 4         return combination(candidates, target, 0);
 5     }
 6
 7     public List<List<Integer>> combination(int[] candidates, int target, int start) {
 8         List<List<Integer>> list = new ArrayList<>();
 9         if(candidates == null || candidates.length == 0) return list;
10
11         for(int i = start; i < candidates.length; i++){
12             if(candidates[i] < target){
13                 List<List<Integer>> tlist = combination(candidates, target - candidates[i], i);
14                 if(tlist.size() > 0){
15                     for(List<Integer> alist : tlist){
16                         alist.add(0, candidates[i]);
17                     }
18                     list.addAll(tlist);
19                 }
20             }
21             else if(candidates[i] == target){
22                 List<Integer> tlist = new LinkedList<>();
23                 tlist.add(target);
24                 list.add(tlist);
25             }
26             else break;
27         }
28         return list;
29     }
30 }
时间: 2024-11-07 00:56:54

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