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39. Combination Sum
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:
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All numbers (including target) will be positive integers.
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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]
]思路分析
参考 46. Permutations 认真学习一下回溯思想。
为了避免重复，每次只遍历不比自己小的元素。（比自己小的元素已经被小元素遍历过了。）
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一刷
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二刷
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三刷
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/**
* Runtime: 164 ms, faster than 5.05% of Java online submissions for Combination Sum.
* Memory Usage: 45.5 MB, less than 5.19% of Java online submissions for Combination Sum.
* <p>
* ↓
* <p>
* Runtime: 9 ms, faster than 17.31% of Java online submissions for Combination Sum.
* Memory Usage: 41 MB, less than 5.19% of Java online submissions for Combination Sum.
*
* @author D瓜哥 · https://www.diguage.com
* @since 2018-09-16 21:56
*/
public List<List<Integer>> combinationSum(int[] candidates, int target) {
if (null == candidates || candidates.length == 0) {
return Collections.emptyList();
}
Arrays.sort(candidates);
List<List<Integer>> result = new ArrayList<>();
backtrack(candidates, target, result, new ArrayList<>());
return result;
}
private void backtrack(int[] candidates, int target, List<List<Integer>> result, List<Integer> current) {
if (target == 0) {
result.add(new ArrayList<>(current));
}
if (target < 0) {
return;
}
for (int candidate : candidates) {
if (!current.isEmpty() && current.get(current.size() - 1) > candidate) {
continue;
}
current.add(candidate);
backtrack(candidates, target - candidate, result, current);
current.remove(current.size() - 1);
}
}
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2024-07-09 19:54:44
*/
public List<List<Integer>> combinationSum(int[] candidates, int target) {
if (candidates == null || candidates.length == 0 || target <= 0) {
return Collections.emptyList();
}
Arrays.sort(candidates);
List<List<Integer>> result = new ArrayList<>();
backtrack(candidates, target, 0, result, new ArrayList<>());
return result;
}
private void backtrack(int[] candidates, int target, int start,
List<List<Integer>> result, List<Integer> perm) {
// 子集和等于 target 时，记录解
if (target == 0) {
result.add(new ArrayList<>(perm));
return;
}
// 剪枝二：从 start 开始遍历，避免生成重复子集
for (int i = start; i < candidates.length; i++) {
int num = candidates[i];
// 剪枝一：若子集和超过 target ，则直接结束循环
if (target < num) {
// 上面排过序，这里直接break
break;
}
perm.add(num);
backtrack(candidates, target - num, i, result, perm);
perm.remove(perm.size() - 1);
}
}
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2024-09-13 22:56:47
*/
public List<List<Integer>> combinationSum(int[] candidates, int target) {
List<List<Integer>> result = new LinkedList<>();
Arrays.sort(candidates);
List<Integer> path = new ArrayList();
backtrack(candidates, target, result, path, 0);
return result;
}
private void backtrack(int[] candidates, int target,
List<List<Integer>> result, List<Integer> path, int index) {
if (target == 0) {
result.add(new ArrayList(path));
return;
}
for (int i = index; i < candidates.length; i++) {
int num = candidates[i];
if (target < num) {
break;
}
path.add(num);
backtrack(candidates, target - num, result, path, i);
path.removeLast();
}
}