题目:
给定一个无重复元素的数组 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。
candidates 中的数字可以无限制重复被选取。
说明：
- 所有数字（包括 target）都是正整数。
- 解集不能包含重复的组合。
示例 1：
- 输入：candidates = [2,3,6,7], target = 7,
- 所求解集为：
[
[7],
[2,2,3]
]
示例 2：
- 输入：candidates = [2,3,5], target = 8,
- 所求解集为：
[
[2,2,2,2],
[2,3,3],
[3,5]
]
java:
public class Leetcode039 {
public static List<List<Integer>> combinationSum(int[] candidates, int target) {
List<List<Integer>> res = new ArrayList<>();
//先进行排序.
Arrays.sort(candidates);
backtrack(res, new ArrayList<>(), candidates, target, 0, 0);
return res;
}
private static void backtrack(List<List<Integer>> res, List<Integer> path, int[] candidates, int target, int sum, int idx) {
//找到满足条件的结果.
if (sum == target) {
res.add(new ArrayList<>(path));
return;
}
for (int i = idx; i < candidates.length; i++) {
if (sum + candidates[i] > target) break;
path.add(candidates[i]);
backtrack(res, path, candidates, target, sum + candidates[i], i);
path.remove(path.size() - 1);
}
}
public static void main(String[] args) {
int[] candidates = {2,3,6,7};
int target = 7;
System.out.println(combinationSum(candidates, target));
}
}
Go:
package LeetCode
import (
"sort"
)
var (
resSort [][]int
pathSort []int
)
func CombinationSum(candidates []int, target int) [][]int {
resSort, pathSort = make([][]int, 0), make([]int, len(candidates))
//排序.
sort.Ints(candidates)
dfsSort(candidates, 0, target)
return resSort
}
func dfsSort(candidates []int, start int, target int) {
if target == 0 {
tmp := make([]int, len(pathSort))
copy(tmp, pathSort)
resSort = append(resSort, tmp)
removeZero(resSort)
return
}
for i := start; i < len(candidates); i++ {
if candidates[i] > target {
break
}
pathSort = append(pathSort, candidates[i])
dfsSort(candidates, i, target-candidates[i])
pathSort = pathSort[:len(pathSort)-1]
}
}
func removeZero(array [][]int) [][]int {
for i := 0; i < len(array); i++ {
slice := array[i]
num := make([]int, 0)
for _, value := range slice {
if value != 0 {
num = append(num, value)
}
}
array[i] = num
}
return array
}
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