Topic
- Array
- Backtracking
Description
https://leetcode.com/problems/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 sum to target.
Each number in candidates may only be used once in the combination.
Note: The solution set must not contain duplicate combinations.
Example 1:
Input: candidates = [10,1,2,7,6,1,5], target = 8
Output:
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]
Example 2:
Input: candidates = [2,5,2,1,2], target = 5
Output:
[
[1,2,2],
[5]
]
Constraints:
- 1 <=
candidates.length<= 100 - 1 <=
candidates[i]<= 50 - 1 <=
target<= 30
Analysis
Submission
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
public class CombinationSumII {
// 版本一：公众号“代码随想录”的
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
List<List<Integer>> result = new ArrayList<>();
List<Integer> path = new ArrayList<>();
Arrays.sort(candidates);
boolean[] used = new boolean[candidates.length];
backtracking(path, candidates, target, used, 0, 0, result);
return result;
}
private void backtracking(List<Integer> path, int[] candidates, int target, //
boolean[] used, int sum, int startIndex, List<List<Integer>> result) {
if (target == sum) {
result.add(new ArrayList<>(path));
return;
}
for (int i = startIndex; i < candidates.length && sum + candidates[i] <= target; i++) {
if (i > 0 && candidates[i - 1] == candidates[i] && used[i - 1] == false)
continue;
sum += candidates[i];
path.add(candidates[i]);
used[i] = true;
backtracking(path, candidates, target, used, sum, i + 1, result);
sum -= candidates[i];
path.remove(path.size() - 1);
used[i] = false;
}
}
// 版本二，跟版本一相比，没有sum，used变量
public List<List<Integer>> combinationSum2_(int[] candidates, int target) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
List<Integer> path = new ArrayList<Integer>();
Arrays.sort(candidates);
backtracking_(candidates, 0, target, path, res);
return res;
}
private void backtracking_(int[] candidates, int startIndex, int target, List<Integer> path,
List<List<Integer>> res) {
if (target == 0) {
res.add(new ArrayList<>(path));
return;
}
for (int i = startIndex; i < candidates.length && target >= candidates[i]; i++) {
if (i > startIndex && candidates[i] == candidates[i - 1])
continue;
path.add(candidates[i]);
backtracking_(candidates, i + 1, target - candidates[i], path, res);
path.remove(path.size() - 1);
}
}
// 版本三，将版本一的used剔除掉，以及backtracking__方法体的 `i > 0 &&` 改为 `i > startIndex &&`
public List<List<Integer>> combinationSum2__(int[] candidates, int target) {
List<List<Integer>> result = new ArrayList<>();
List<Integer> path = new ArrayList<>();
Arrays.sort(candidates);
backtracking__(path, candidates, target, 0, 0, result);
return result;
}
private void backtracking__(List<Integer> path, int[] candidates, int target, //
int sum, int startIndex, List<List<Integer>> result) {
if (target == sum) {
result.add(new ArrayList<>(path));
return;
}
for (int i = startIndex; i < candidates.length && sum + candidates[i] <= target; i++) {
if (i > startIndex && candidates[i - 1] == candidates[i])
continue;
sum += candidates[i];
path.add(candidates[i]);
backtracking__(path, candidates, target, sum, i + 1, result);
sum -= candidates[i];
path.remove(path.size() - 1);
}
}
}
Test
import static org.junit.Assert.*;
import java.util.Arrays;
import org.hamcrest.collection.IsIterableContainingInAnyOrder;
import org.junit.Test;
public class CombinationSumIITest {
@Test
@SuppressWarnings("unchecked")
public void test() {
CombinationSumII obj = new CombinationSumII();
assertThat(obj.combinationSum2(new int[] {
10, 1, 2, 7, 6, 1, 5 }, 8), //
IsIterableContainingInAnyOrder.containsInAnyOrder(Arrays.asList(1, 1, 6), Arrays.asList(1, 2, 5), //
Arrays.asList(1, 7), Arrays.asList(2, 6)));
assertThat(obj.combinationSum2(new int[] {
2, 5, 2, 1, 2 }, 5), //
IsIterableContainingInAnyOrder.containsInAnyOrder(Arrays.asList(1, 2, 2), Arrays.asList(5)));
}
@Test
@SuppressWarnings("unchecked")
public void test2() {
CombinationSumII obj = new CombinationSumII();
assertThat(obj.combinationSum2_(new int[] {
10, 1, 2, 7, 6, 1, 5 }, 8), //
IsIterableContainingInAnyOrder.containsInAnyOrder(Arrays.asList(1, 1, 6), Arrays.asList(1, 2, 5), //
Arrays.asList(1, 7), Arrays.asList(2, 6)));
assertThat(obj.combinationSum2_(new int[] {
2, 5, 2, 1, 2 }, 5), //
IsIterableContainingInAnyOrder.containsInAnyOrder(Arrays.asList(1, 2, 2), Arrays.asList(5)));
assertThat(obj.combinationSum2_(new int[] {
1, 1, 1}, 3), //
IsIterableContainingInAnyOrder.containsInAnyOrder(Arrays.asList(1, 1, 1)));
}
@Test
@SuppressWarnings("unchecked")
public void test3() {
CombinationSumII obj = new CombinationSumII();
assertThat(obj.combinationSum2__(new int[] {
10, 1, 2, 7, 6, 1, 5 }, 8), //
IsIterableContainingInAnyOrder.containsInAnyOrder(Arrays.asList(1, 1, 6), Arrays.asList(1, 2, 5), //
Arrays.asList(1, 7), Arrays.asList(2, 6)));
assertThat(obj.combinationSum2__(new int[] {
2, 5, 2, 1, 2 }, 5), //
IsIterableContainingInAnyOrder.containsInAnyOrder(Arrays.asList(1, 2, 2), Arrays.asList(5)));
assertThat(obj.combinationSum2__(new int[] {
1, 1, 1}, 3), //
IsIterableContainingInAnyOrder.containsInAnyOrder(Arrays.asList(1, 1, 1)));
}
}