Subsets II
描述
Given a collection of integers that might contain duplicates, S, return all possible subsets.
Note:
Elements in a subset must be in non-descending order.
The solution set must not contain duplicate subsets.
For example,
If S = [1,2,2], a solution is:
[
[2],
[1],
[1,2,2],
[2,2],
[1,2],
[]
]
分析
这题有重复元素，但本质上，跟上一题很类似，上一题中元素没有重复，相当于每个元素只能选0或1次，这里扩充到了每个元素可以选0到若干次而已。
递归
增量构造法
// Subsets II
// 增量构造法，版本1，时间复杂度O(2^n)，空间复杂度O(n)
public class Solution {
public List<List<Integer>> subsetsWithDup(int[] nums) {
Arrays.sort(nums); // 必须排序
List<List<Integer>> result = new ArrayList<>();
List<Integer> path = new ArrayList<>();
dfs(nums, 0, path, result);
return result;
}
private static void dfs(int[] nums, int start, List<Integer> path,
List<List<Integer>> result) {
result.add(new ArrayList<Integer>(path));
for (int i = start; i < nums.length; i++) {
if (i != start && nums[i] == nums[i-1]) continue;
path.add(nums[i]);
dfs(nums, i + 1, path, result);
path.remove(path.size() - 1);
}
}
}
// Subsets II
// 增量构造法，版本2，时间复杂度O(2^n)，空间复杂度O(n)
public class Solution {
public List<List<Integer>> subsetsWithDup(int[] nums) {
Arrays.sort(nums); // 必须排序
List<List<Integer>> result = new ArrayList<>();
List<Integer> path = new ArrayList<>(); // 中间结果
// 记录每个元素的出现次数
HashMap<Integer, Integer> counterMap = new HashMap<>();
for (int i : nums) {
counterMap.put(i, counterMap.getOrDefault(i, 0) + 1);
}
// 将HashMap里的pair拷贝到一个数组里
Pair[] counters = new Pair[counterMap.size()];
int i = 0;
for (Map.Entry<Integer, Integer> entry : counterMap.entrySet()) {
counters[i++] = new Pair(entry.getKey(), entry.getValue());
}
Arrays.sort(counters);
dfs(counters, 0, path, result);
return result;
}
private static void dfs(Pair[] counters, int step, List<Integer> path,
List<List<Integer>> result) {
if (step == counters.length) {
result.add(new ArrayList<>(path));
return;
}
for (int i = 0; i <= counters[step].value; i++) {
for (int j = 0; j < i; ++j) {
path.add(counters[step].key);
}
dfs(counters, step + 1, path, result);
for (int j = 0; j < i; ++j) {
path.remove(path.size() - 1);
}
}
}
static class Pair implements Comparable<Pair> {
int key;
int value;
public Pair(int key, int value) {
this.key = key;
this.value = value;
}
@Override
public int compareTo(Pair o) {
if (this.key < o.key) return -1;
else if (this.key > o.key) return 1;
else {
return this.value - o.value;
}
}
}
}
位向量法
// Subsets II
// 位向量法，时间复杂度O(2^n)，空间复杂度O(n)
public class Solution {
public List<List<Integer>> subsetsWithDup(int[] nums) {
Arrays.sort(nums); // 必须排序
List<List<Integer>> result = new ArrayList<>();
// 记录每个元素的出现次数
HashMap<Integer, Integer> counterMap = new HashMap<>();
for (int i : nums) {
counterMap.put(i, counterMap.getOrDefault(i, 0) + 1);
}
// 将HashMap里的pair拷贝到一个数组里
Pair[] counters = new Pair[counterMap.size()];
int i = 0;
for (Map.Entry<Integer, Integer> entry : counterMap.entrySet()) {
counters[i++] = new Pair(entry.getKey(), entry.getValue());
}
Arrays.sort(counters);
// 每个元素选择了多少个
HashMap<Integer, Integer> selected = new HashMap<>();
for (Pair p : counters) {
selected.put(p.key, 0 );
}
dfs(nums, counters, selected, 0, result);
return result;
}
private static void dfs(int[] nums, Pair[] counters, HashMap<Integer, Integer> selected,
int step, List<List<Integer>> result) {
if (step == counters.length) {
ArrayList<Integer> subset = new ArrayList<>();
for (Pair p : counters) {
for (int i = 0; i < selected.get(p.key); ++i) {
subset.add(p.key);
}
}
result.add(subset);
return;
}
for (int i = 0; i <= counters[step].value; i++) {
selected.put(counters[step].key, i);
dfs(nums, counters, selected, step + 1, result);
}
}
static class Pair implements Comparable<Pair> {
int key;
int value;
public Pair(int key, int value) {
this.key = key;
this.value = value;
}
@Override
public int compareTo(Pair o) {
if (this.key < o.key) return -1;
else if (this.key > o.key) return 1;
else {
return this.value - o.value;
}
}
}
}
迭代
增量构造法
// Subsets II
// 增量构造法
// 时间复杂度O(2^n)，空间复杂度O(1)
public class Solution {
public List<List<Integer>> subsetsWithDup(int[] nums) {
Arrays.sort(nums); // 必须排序
List<List<Integer>> result = new ArrayList<>();
result.add(new ArrayList<Integer>());
int previous_size = 0;
for (int i = 0; i < nums.length; ++i) {
final int size = result.size();
for (int j = 0; j < size; ++j) {
if (i == 0 || nums[i] != nums[i-1] || j >= previous_size) {
result.add(new ArrayList<>(result.get(j)));
result.get(result.size() - 1).add(nums[i]);
}
}
previous_size = size;
}
return result;
}
}
二进制法
// Subsets II
// 二进制法，时间复杂度O(2^n)，空间复杂度O(1)
public class Solution {
public List<List<Integer>> subsetsWithDup(int[] nums) {
Arrays.sort(nums); // 必须排序
// 用 set 去重，不能用 unordered_set，因为输出要求有序
LinkedHashSet<ArrayList<Integer>> result = new LinkedHashSet<>();
final int n = nums.length;
ArrayList<Integer> v = new ArrayList<>();
for (int i = 0; i < 1 << n; ++i) {
for (int j = 0; j < n; ++j) {
if ((i & 1 << j) > 0)
v.add(nums[j]);
}
result.add(new ArrayList<>(v));
v.clear();
}
List<List<Integer>> realResult = new ArrayList<>();
for (ArrayList<Integer> list : result) {
realResult.add(list);
}
return realResult;
}
}