46题目描述:
给定一个不含重复数字的数组 nums ,返回其 所有可能的全排列 。你可以 按任意顺序 返回答案。
解析:
回溯法
Java:
class Solution {
public List<List<Integer>> permute(int[] nums) {
List<List<Integer>> res = new ArrayList<>();
List<Integer> list = new ArrayList<>();
backtrack(res, list, nums);
return res;
}
public void backtrack(List<List<Integer>> res, List<Integer> list, int[] nums) {
if(list.size() == nums.length) {
res.add(new ArrayList<Integer>(list));
return;
}
for(int i = 0; i < nums.length; i++) {
if(!list.contains(nums[i])) {
list.add(nums[i]);
backtrack(res, list, nums);
list.remove(list.size() - 1);
}
}
}
}
JavaScript:
var permute = function(nums) {
const res = [], path = [];
backtracking(nums, nums.length, []);
return res;
function backtracking(n, k, used) {
if(path.length === k) {
res.push(Array.from(path));
return;
}
for (let i = 0; i < k; i++ ) {
if(used[i]) continue;
path.push(n[i]);
used[i] = true; // 同支
backtracking(n, k, used);
path.pop();
used[i] = false;
}
}
};
47题目描述:
给定一个可包含重复数字的序列 nums ,按任意顺序 返回所有不重复的全排列。
解析:
回溯法
Java:
class Solution {
public List<List<Integer>> permuteUnique(int[] nums) {
List<List<Integer>> res = new ArrayList<>();
if(nums == null || nums.length ==0) {
return res;
}
int[] visited = new int[nums.length];
Arrays.sort(nums);
backTrack(res, nums, new ArrayList<Integer>(), visited);
return res;
}
private void backTrack(List<List<Integer>> res, int[] nums, List<Integer> list, int[] visited) {
if(list.size() == nums.length) {
res.add(new ArrayList<>(list));
return;
}
for(int i = 0; i < nums.length; i++) {
if(visited[i] == 1) {
continue;
}
if(i > 0 && nums[i] == nums[i - 1] && visited[i - 1] == 0) {
continue;
}
list.add(nums[i]);
visited[i] = 1;
backTrack(res, nums, list, visited);
visited[i] = 0;
list.remove(list.size() - 1);
}
}
}
JavaScript:
var permuteUnique = function(nums) {
const res = [], list = [];
if(nums == null || nums.length ==0) {
return res;
}
nums = nums.sort((a, b) => a - b);
dfs(nums, nums.length, []);
return res;
function dfs(n, k, visited) {
if(list.length === k) {
res.push(Array.from(list));
return;
}
for(let i = 0; i < k; i++) {
if(visited[i]) {
continue;
}
if(i > 0 && nums[i] == nums[i - 1] && visited[i - 1] == 0) {
continue;
}
list.push(n[i]);
visited[i] = true;
dfs(n, k, visited);
list.pop();
visited[i] = false;
}
}
};