题目
地址:https://leetcode.com/problems/subsets/
Given a set of distinct integers, nums, return all possible subsets (the power set).
Note: The solution set must not contain duplicate subsets.
Example:
Input: nums = [1,2,3]
Output:
[
[3],
[1],
[2],
[1,2,3],
[1,3],
[2,3],
[1,2],
[]
]
1. DFS深度优先回溯解法
思路解析:
- 遍历数组中的元素,要么选择,要么不选择。
- 注意退出条件即可
if (nums == null || index == nums.length)
。表示可能数组为空,可能index已经越界,则添加到结果列表中。
package backtracking;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
// https://leetcode.com/problems/subsets/
public class Subsets {
public static void main(String[] args) {
int[] nums = new int[]{1,2,3};
Subsets obj = new Subsets();
List<List<Integer>> resultList = obj.subsets(nums);
System.out.println(Arrays.toString(resultList.toArray()));
}
public List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> resultList = new ArrayList<List<Integer>>();
// dfs
dfs(nums, resultList, new ArrayList<Integer>(), 0);
return resultList;
}
private void dfs(int[] nums, List<List<Integer>> resultList, List<Integer> list, int index) {
// exit
if (nums == null || index == nums.length) {
resultList.add(new ArrayList<Integer>(list));
return;
}
// add item
list.add(nums[index]);
dfs(nums, resultList, list, index + 1);
// not add item
list.remove(list.size() - 1);
dfs(nums, resultList, list, index + 1);
}
}
2. 遍历执行
从一个空列表List<List<Integer>> outputList
开始,添加一个空列表new ArrayList<Integer>()
,遍历所有数据(比如:{1, 2, 3}
),
- 新建一个空列表
List<List<Integer>> newList
, - 遍历列表
List<List<Integer>> outputList
,把已有子项都添加上新的数字,
比如遍历到1,
[ ] > [1]
比如遍历到2,
[ ] > [2]
[1] > [1, 2]
- 遍历列表
new ArrayList<Integer>()
,把新生成的子列表追加到已有列表的后面List<List<Integer>> outputList
。
public List<List<Integer>> subsetsWithRecursion(int[] nums) {
List<List<Integer>> outputList = new ArrayList<List<Integer>>();
outputList.add(new ArrayList<Integer>());
for (int num: nums) {
List<List<Integer>> newList = new ArrayList<List<Integer>>();
for (List<Integer> list: outputList) {
newList.add(new ArrayList<Integer>(list) {{ add(num); }});
}
for (List<Integer> list: newList) {
outputList.add(list);
}
}
return outputList;
}
3. 回溯解法,达到指定长度就返回
public List<List<Integer>> subsetsWithBacktrack(int[] nums) {
List<List<Integer>> resultList = new ArrayList<List<Integer>>();
for (int len = 0; len <= nums.length; len++) {
// backtrack
backtrack(nums, resultList, new ArrayList<Integer>(), 0, len);
}
return resultList;
}
private void backtrack(int[] nums, List<List<Integer>> resultList, List<Integer> list, int first, int len) {
// exit
if (list.size() == len) {
resultList.add(new ArrayList<Integer>(list));
return;
}
if (first == nums.length) {
return;
}
list.add(nums[first]);
backtrack(nums, resultList, list, first + 1, len);
list.remove(list.size() - 1);
backtrack(nums, resultList, list, first + 1, len);
}
4. 二进制位组装
算法思想来自 Donald E.Knuth. 组装成跟数组长度一样的二进制,如果为1则添加进来,否则就不添加。
public List<List<Integer>> subsetsWithBinarySorted(int[] nums) {
List<List<Integer>> resultList = new ArrayList<List<Integer>>();
int n = nums.length;
for (int i = (int)Math.pow(2, n); i < (int)Math.pow(2, n + 1); i++) {
// generate bitmask, from 0..00 to 1..11
String bitmask = Integer.toBinaryString(i).substring(1);
// append subset corresponding to that bitmask
List<Integer> list = new ArrayList<Integer>();
for (int k = 0; k < n; k++) {
if (bitmask.charAt(k) == '1') {
list.add(nums[k]);
}
}
resultList.add(list);
}
return resultList;
}
代码下载
参考
https://leetcode.com/problems/subsets/solution/
https://www-cs-faculty.stanford.edu/~knuth/taocp.html