LeetCode-全排列

46. Permutations

Given a collection of distinct numbers, return all possible permutations.

For example,
[1,2,3] have the following permutations:

[  [1,2,3],  [1,3,2],  [2,1,3],  [2,3,1],  [3,1,2],  [3,2,1]]

求全排列，DFS的典型应用

package solutions._46;/** * 46. Permutations */import java.util.ArrayList;import java.util.LinkedList;import java.util.List;class Solution {    private boolean visited[];    private List<List<Integer>> result;    private List<Integer> cur;    private void DFS(int[] nums, int len) {        if (len == nums.length) {            result.add(new ArrayList<>(cur));            return;        }        for (int i = 0; i < nums.length; i++) {            if (!visited[i]) {                visited[i] = true;                cur.add(nums[i]);                DFS(nums, len + 1);                visited[i] = false;                cur.remove(cur.size() - 1);            }        }    }    public List<List<Integer>> permute(int[] nums) {        visited = new boolean[nums.length];        result = new LinkedList<>();        cur = new LinkedList<>();        DFS(nums, 0);        return result;    }    public static void main(String[] args) {        Solution solution = new Solution();        int[] arr = {1, 2};        List<List<Integer>> lists = solution.permute(arr);        System.out.println(lists);    }}

77. Combinations

Given two integers n and k, return all possible combinations of k numbers out of 1 … n.

For example,
If n = 4 and k = 2, a solution is:

[  [2,4],  [3,4],  [2,3],  [1,2],  [1,3],  [1,4],]

基本类似于全排列，只不过有顺序性从而避免重复（[1,4]和[4,1]一样）。

package solutions._77;/** * 77. Combinations */import java.util.ArrayList;import java.util.LinkedList;import java.util.List;class Solution {    private boolean[] visited;    private List<List<Integer>> result;    private List<Integer> cur;    private int n;    private int k;    private void DFS(int pos, int len) {        if (len == k) {            result.add(new ArrayList<>(cur));            return;        }        for (int i = pos; i <= n; i++) {            if (!visited[i]) {                visited[i] = true;                cur.add(i);                DFS(i, len + 1);                cur.remove(cur.size() - 1);                visited[i] = false;            }        }    }    public List<List<Integer>> combine(int n, int k) {        visited = new boolean[n + 1];        cur = new LinkedList<>();        result = new LinkedList<>();        this.n = n;        this.k = k;        DFS(1, 0);        return result;    }    public static void main(String[] args) {        Solution solution = new Solution();        List<List<Integer>> lists = solution.combine(4, 2);        System.out.println(lists);    }}

47. Permutations II

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

For example,
[1,1,2] have the following unique permutations:

[  [1,1,2],  [1,2,1],  [2,1,1]]

可以将结果集保存在set中，有点投机取巧了。

package solutions._47;/** * 47. Permutations II */import java.util.*;class Solution {    private boolean visited[];    private Set<List<Integer>> sets;    private List<Integer> cur;    private void DFS(int[] nums, int len) {        if (len == nums.length) {            sets.add(new ArrayList<>(cur));            return;        }        for (int i = 0; i < nums.length; i++) {            if (!visited[i]) {                visited[i] = true;                cur.add(nums[i]);                DFS(nums, len + 1);                visited[i] = false;                cur.remove(cur.size() - 1);            }        }    }    public List<List<Integer>> permuteUnique(int[] nums) {        visited = new boolean[nums.length];        sets = new HashSet<>();        cur = new LinkedList<>();        DFS(nums, 0);        List<List<Integer>> result = new LinkedList<>();        result.addAll(sets);        return result;    }    public static void main(String[] args) {        Solution solution = new Solution();        int[] arr = {1, 1, 0, 0, 1, -1, -1, 1};        List<List<Integer>> lists = solution.permuteUnique(arr);        System.out.println(lists);    }}