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40. Combination Sum II
Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
Each number in candidates may only be used once in the combination.
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 = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]Example 2:
Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
[1,2,2],
[5]
]思路分析
这道题的关键是由于候选值不能重复使用，所以需要向下传递起始位置。可以对比一下 39. Combination Sum 的处理上的不同之处。
思考一下解决重复值时是怎么剪枝的？
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一刷
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二刷
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三刷
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/**
* Runtime: 2 ms, faster than 100.00% of Java online submissions for Combination Sum II.
* Memory Usage: 39.5 MB, less than 54.74% of Java online submissions for Combination Sum II.
*
* @author D瓜哥 · https://www.diguage.com
* @since 2020-01-27 19:20
*/
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
if (Objects.isNull(candidates) || candidates.length == 0) {
return Collections.emptyList();
}
Arrays.sort(candidates);
List<List<Integer>> result = new ArrayList<>();
backtrack(candidates, 0, target, result, new ArrayDeque<>());
return new ArrayList<>(result);
}
private void backtrack(int[] candidates, int beginIndex, int target,
List<List<Integer>> result, Deque<Integer> current) {
if (target == 0) {
result.add(new ArrayList<>(current));
return;
}
for (int i = beginIndex; i < candidates.length; i++) {
int candidate = candidates[i];
if (target < candidate) {
break;
}
if (!current.isEmpty() && current.peekLast() > candidate) {
continue;
}
if (beginIndex < i && candidate == candidates[i - 1]) {
continue;
}
current.addLast(candidate);
// 因为元素不可以重复使用，这里递归传递下去的是 i + 1 而不是 i
backtrack(candidates, i + 1, target - candidate, result, current);
current.removeLast();
}
}
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2024-07-10 16:52:25
*/
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
List<List<Integer>> result = new ArrayList<>();
Arrays.sort(candidates);
backtrack(candidates, result, 0, target, new ArrayList<>());
return result;
}
private void backtrack(int[] candidates, List<List<Integer>> result,
int start, int target, List<Integer> path) {
// 子集和等于 target 时，记录解
if (target == 0) {
result.add(new ArrayList<>(path));
return;
}
// 剪枝二：从 start 开始遍历，避免生成重复子集
// 剪枝三：从 start 开始遍历，避免重复选择同一元素
for (int i = start; i < candidates.length; i++) {
// 剪枝一：若子集和超过 target ，则直接结束循环
// 这是因为数组已排序，后边元素更大，子集和一定超过 target
if (target < candidates[i]) {
break;
}
// 剪枝四：如果该元素与左边元素相等，说明该搜索分支重复，直接跳过
if (start < i && candidates[i - 1] == candidates[i]) {
continue;
}
path.add(candidates[i]);
// 进行下一轮选择
backtrack(candidates, result, i + 1, target - candidates[i], path);
path.remove(path.size() - 1);
}
}
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2024-09-16 17:31:07
*/
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
Arrays.sort(candidates);
List<List<Integer>> result = new ArrayList<>();
backtrack(candidates, target, result, 0, new ArrayList<>());
return result;
}
private void backtrack(int[] candidates, int target,
List<List<Integer>> result, int start, List<Integer> path) {
if (target == 0) {
result.add(new ArrayList(path));
return;
}
for (int i = start; i < candidates.length; i++) {
int num = candidates[i];
if (target < num) {
break;
}
if (start < i && candidates[i - 1] == candidates[i]) {
continue;
}
path.add(num);
backtrack(candidates, target - num, result, i + 1, path);
path.remove(path.size() - 1);
}
}