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# 39.Combination-Sum

## 题目描述

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]
]

## 代码

### Approach 1: backtracking

class Solution {
public List<List<Integer>> combinationSum(int[] candidates, int target) {
List<List<Integer>> result = new ArrayList<>();
if (candidates == null) return result;
List<Integer> combination = new ArrayList<>();
Arrays.sort(candidates); // sort
dfs(candidates, 0, target, combination, result);
return result;
}
private void dfs(int[] candidates, int index, int target, List<Integer> combination, List<List<Integer>> result) {
if (target == 0) {
return;
}
for (int i = index; i < candidates.length; i++) {
if (candidates[i] > target) {
break;
}
// 去重
// if (i != 0 && candidates[i] == candidates[i - 1]) {
// continue;
// }
dfs(candidates, i, target - candidates[i], combination, result);
combination.remove(combination.size() - 1);
}
}
}

### Approach #2 Reuse

public class Solution {
public List<List<Integer>> combinationSum(int[] candidates, int target) {
List<List<Integer>> result = new ArrayList<>();
if (candidates == null) return result;
dfs(candidates, 0, target, new ArrayList<Integer>(), result);
return result;
}
private void dfs(int[] candidates, int index, int target, List<Integer> combination, List<List<Integer>> result) {
if (target == 0) {
return;
}
if (index == candidates.length || target < 0) {
return;
}
dfs(candidates, index + 1, target, combination, result);
if (index > 0 && candidates[index] == candidates[index - 1]) {
return;
}