Difficulty:: Medium
Given a collection of distinct integers, return all possible permutations.
Example:
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| Input: [1,2,3]
Output:
[
[1,2,3],
[1,3,2],
[2,1,3],
[2,3,1],
[3,1,2],
[3,2,1]
]
|
Solution
Language: Java
简单做法，66%
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| class Solution {
public List<List<Integer>> permute(int[] nums) {
List<List<Integer>> result = new ArrayList<>();
if (nums == null) {
return result;
}
dfsHelper(nums, result, new ArrayList<Integer>());
return result;
}
private void dfsHelper(int[] nums, List<List<Integer>> result, List<Integer> p) {
if (p.size() == nums.length) {
result.add(new ArrayList<Integer>(p));
return;
}
for (int i = 0; i < nums.length; i++) {
if (p.contains(nums[i])) {
continue;
}
p.add(nums[i]);
dfsHelper(nums, result, p);
p.remove(p.size() - 1);
}
}
}
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记忆化搜索 100%
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| class Solution {
public List<List<Integer>> permute(int[] nums) {
List<List<Integer>> result = new ArrayList<>();
if (nums == null) {
return result;
}
boolean[] visited = new boolean[nums.length];
dfsHelper(nums, result, new ArrayList<Integer>(), visited);
return result;
}
private void dfsHelper(int[] nums, List<List<Integer>> result, List<Integer> p, boolean[] visited) {
if (p.size() == nums.length) {
result.add(new ArrayList<Integer>(p));
return;
}
for (int i = 0; i < nums.length; i++) {
if (visited[i]) {
continue;
}
p.add(nums[i]);
visited[i] = true;
dfsHelper(nums, result, p, visited);
visited[i] = false;
p.remove(p.size() - 1);
}
}
}
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