组合总和
力扣链接
题目描述
给定一个无重复元素的正整数数组 candidates
和一个正整数 target
,找出 candidates
中所有可以使数字和为目标数 target
的唯一组合。
candidates
中的数字可以无限制重复被选取。如果至少一个所选数字数量不同,则两种组合是唯一的。
对于给定的输入,保证和为 target
的唯一组合数少于 150
个。
示例 1:
输入: candidates = [2,3,6,7], target = 7
输出: [[7],[2,2,3]]
示例 2:
输入: candidates = [2,3,5], target = 8
输出: [[2,2,2,2],[2,3,3],[3,5]]
示例 3:
输入: candidates = [2], target = 1
输出: []
示例 4:
输入: candidates = [1], target = 1
输出: [[1]]
示例 5:
输入: candidates = [1], target = 2
输出: [[1,1]]
提示:
1 <= candidates.length <= 30
1 <= candidates[i] <= 200
candidate
中的每个元素都是独一无二的。1 <= target <= 500
解题思路
- dfs + 剪枝
代码
public List<List<Integer>> combinationSum(int[] candidates, int target) {
int size = candidates.length;
List<List<Integer>> list = new ArrayList<>();
Deque<Integer> path = new LinkedList<>();
dfs(candidates, 0, size, target, list, path);
return list;
}
void dfs(int[] candidates, int begin, int size, int target, List<List<Integer>> list, Deque<Integer> path) {
if (target < 0) {
return;
}
if (target == 0) {
list.add(new ArrayList<>(path));
}
for (int i = begin; i < size; i++) {
path.add(candidates[i]);
dfs(candidates, i, size, target - candidates[i], list, path);
path.removeLast();
}
}
复杂度
代码(dfs+剪枝)
public class Solution {
public List<List<Integer>> combinationSum(int[] candidates, int target) {
int size = candidates.length;
List<List<Integer>> list = new ArrayList<>();
Deque<Integer> path = new LinkedList<>();
// 排序是剪枝的前提
Arrays.sort(candidates);
dfs(candidates, 0, size, target, list, path);
return list;
}
void dfs(int[] candidates, int begin, int size, int target, List<List<Integer>> list, Deque<Integer> path) {
if (target == 0) {
list.add(new ArrayList<>(path));
}
for (int i = begin; i < size; i++) {
//剪枝
int number = target - candidates[i];
if (number < 0) {
break;
}
path.add(candidates[i]);
dfs(candidates, i, size, number, list, path);
path.removeLast();
}
}
}