Permutations
描述
Given a collection of 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], and [3,2,1].
next_permutation()
函数 next_permutation()的具体实现见这节 Next Permutation。
// Permutations
// 重新实现 next_permutation()
// 时间复杂度O(n!)，空间复杂度O(1)
public class Solution {
public List<List<Integer>> permute(int[] nums) {
List<List<Integer>> result = new ArrayList<>();
Arrays.sort(nums);
do {
ArrayList<Integer> one = new ArrayList<>();
for (int i : nums) {
one.add(i);
}
result.add(one);
// 调用的是 2.1.12 节的 next_permutation()
// 而不是 std::next_permutation()
} while(nextPermutation(nums, 0, nums.length));
return result;
}
// 代码来自 2.1.12 节的 next_permutation()
private static boolean nextPermutation(int[] nums, int begin, int end) {
// From right to left, find the first digit(partitionNumber)
// which violates the increase trend
int p = end - 2;
while (p > -1 && nums[p] >= nums[p + 1]) --p;
// If not found, which means current sequence is already the largest
// permutation, then rearrange to the first permutation and return false
if(p == -1) {
reverse(nums, begin, end);
return false;
}
// From right to left, find the first digit which is greater
// than the partition number, call it changeNumber
int c = end - 1;
while (c > 0 && nums[c] <= nums[p]) --c;
// Swap the partitionNumber and changeNumber
swap(nums, p, c);
// Reverse all the digits on the right of partitionNumber
reverse(nums, p+1, end);
return true;
}
private static void swap(int[] nums, int i, int j) {
int tmp = nums[i];
nums[i] = nums[j];
nums[j] = tmp;
}
private static void reverse(int[] nums, int begin, int end) {
end--;
while (begin < end) {
swap(nums, begin++, end--);
}
}
}
递归
本题是求路径本身，求所有解，函数参数需要标记当前走到了哪步，还需要中间结果的引用，最终结果的引用。
扩展节点，每次从左到右，选一个没有出现过的元素。
本题不需要判重，因为状态装换图是一颗有层次的树。收敛条件是当前走到了最后一个元素。
代码
// Permutations
// 深搜，增量构造法
// 时间复杂度O(n!)，空间复杂度O(n)
public class Solution {
public List<List<Integer>> permute(int[] nums) {
Arrays.sort(nums);
List<List<Integer>> result = new ArrayList<>();
List<Integer> path = new ArrayList<>(); // 中间结果
dfs(nums, path, result);
return result;
}
private static void dfs(int[] nums, List<Integer> path,
List<List<Integer>> result) {
if (path.size() == nums.length) { // 收敛条件
result.add(new ArrayList<Integer>(path));
return;
}
// 扩展状态
for (int i : nums) {
// 查找 i 是否在path 中出现过
int pos = path.indexOf(i);
if (pos == -1) {
path.add(i);
dfs(nums, path, result);
path.remove(path.size() - 1);
}
}
}
}