# Combination Sum I & II & IIIleetcode

lpy1990

## Combination 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 (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]

### 回溯法

#### 代码

``````public List<List<Integer>> combinationSum(int[] candidates, int target) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
if (candidates == null || candidates.length == 0 ) {
return res;
}
Arrays.sort(candidates);
List<Integer> tem = new ArrayList<Integer>();
helper(res, tem, target, candidates, 0);
return res;
}
public void helper(List<List<Integer>> res, List<Integer> tem, int target, int[] candidates, int pos) {
if (target == 0) {
return;
}
if (target < 0) {
return;
}
for (int i = pos; i < candidates.length; i++) {
helper(res, tem, target - candidates[i], candidates, i);
tem.remove(tem.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 (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] ``````

### 说明

#### 代码

``````public List<List<Integer>> combinationSum2(int[] candidates, int target) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
if (candidates == null || candidates.length == 0 ) {
return res;
}
Arrays.sort(candidates);
List<Integer> tem = new ArrayList<Integer>();
helper(res, tem, target, candidates, 0);
return res;
}
public void helper(List<List<Integer>> res, List<Integer> tem, int target, int[] candidates, int pos) {
if (target == 0) {
return;
}
if (target < 0) {
return;
}
for (int i = pos; i < candidates.length; i++) {
if (i != pos && candidates[i] == candidates[i - 1]) {
continue;
}
helper(res, tem, target - candidates[i], candidates, i + 1);
tem.remove(tem.size() - 1);
}
}``````

## Combination Sum III

Find all possible combinations of k numbers that add up to a number n,
given that only numbers from 1 to 9 can be used and each combination
should be a unique set of numbers.

Ensure that numbers within the set are sorted in ascending order.

Example 1:

Input: k = 3, n = 7

Output:

[[1,2,4]]

Example 2:

Input: k = 3, n = 9

Output:

[[1,2,6], [1,3,5], [2,3,4]]

### 回溯法

#### 代码

``````public List<List<Integer>> combinationSum3(int k, int n) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
if (n <= 0 || k <= 0) {
return res;
}
List<Integer> tem = new ArrayList<Integer>();
helper(tem, res, 1, n, k);
return res;
}
public void helper(List<Integer> tem, List<List<Integer>> res, int index, int n, int k) {
if (tem.size() == k && n == 0) {
return;
}
if (tem.size() > k || n < 0) {
return;
}
for (int i = index; i <= 9; i++) {