|
| 1 | +--- |
| 2 | +comments: true |
| 3 | +difficulty: 中等 |
| 4 | +edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3437.Permutations%20III/README.md |
| 5 | +--- |
| 6 | + |
| 7 | +<!-- problem:start --> |
| 8 | + |
| 9 | +# [3437. Permutations III 🔒](https://leetcode.cn/problems/permutations-iii) |
| 10 | + |
| 11 | +[English Version](/solution/3400-3499/3437.Permutations%20III/README_EN.md) |
| 12 | + |
| 13 | +## 题目描述 |
| 14 | + |
| 15 | +<!-- description:start --> |
| 16 | + |
| 17 | +<p>Given an integer <code>n</code>, an <strong>alternating permutation</strong> is a permutation of the first <code>n</code> positive integers such that no <strong>two</strong> adjacent elements are <strong>both</strong> odd or <strong>both</strong> even.</p> |
| 18 | + |
| 19 | +<p>Return <em>all such </em><strong>alternating permutations</strong> sorted in lexicographical order.</p> |
| 20 | + |
| 21 | +<p> </p> |
| 22 | +<p><strong class="example">Example 1:</strong></p> |
| 23 | + |
| 24 | +<div class="example-block"> |
| 25 | +<p><strong>Input:</strong> <span class="example-io">n = 4</span></p> |
| 26 | + |
| 27 | +<p><strong>Output:</strong> <span class="example-io">[[1,2,3,4],[1,4,3,2],[2,1,4,3],[2,3,4,1],[3,2,1,4],[3,4,1,2],[4,1,2,3],[4,3,2,1]]</span></p> |
| 28 | +</div> |
| 29 | + |
| 30 | +<p><strong class="example">Example 2:</strong></p> |
| 31 | + |
| 32 | +<div class="example-block"> |
| 33 | +<p><strong>Input:</strong> <span class="example-io">n = 2</span></p> |
| 34 | + |
| 35 | +<p><strong>Output:</strong> <span class="example-io">[[1,2],[2,1]]</span></p> |
| 36 | +</div> |
| 37 | + |
| 38 | +<p><strong class="example">Example 3:</strong></p> |
| 39 | + |
| 40 | +<div class="example-block"> |
| 41 | +<p><strong>Input:</strong> <span class="example-io">n = 3</span></p> |
| 42 | + |
| 43 | +<p><strong>Output:</strong> <span class="example-io">[[1,2,3],[3,2,1]]</span></p> |
| 44 | +</div> |
| 45 | + |
| 46 | +<p> </p> |
| 47 | +<p><strong>Constraints:</strong></p> |
| 48 | + |
| 49 | +<ul> |
| 50 | + <li><code>1 <= n <= 10</code></li> |
| 51 | +</ul> |
| 52 | + |
| 53 | +<!-- description:end --> |
| 54 | + |
| 55 | +## 解法 |
| 56 | + |
| 57 | +<!-- solution:start --> |
| 58 | + |
| 59 | +### 方法一:回溯 |
| 60 | + |
| 61 | +我们设计一个函数 $\textit{dfs}(i),ドル表示当前要填第 $i$ 个位置的数,位置编号从 0ドル$ 开始。 |
| 62 | + |
| 63 | +在 $\textit{dfs}(i)$ 中,如果 $i \geq n,ドル说明所有位置都已经填完,将当前排列加入答案数组中。 |
| 64 | + |
| 65 | +否则,我们枚举当前位置可以填的数 $j,ドル如果 $j$ 没有被使用过,并且 $j$ 和当前排列的最后一个数不同奇偶性,我们就可以将 $j$ 放在当前位置,继续递归填下一个位置。 |
| 66 | + |
| 67 | +时间复杂度 $O(n \times n!),ドル空间复杂度 $O(n)$。其中 $n$ 为排列的长度。 |
| 68 | + |
| 69 | +<!-- tabs:start --> |
| 70 | + |
| 71 | +#### Python3 |
| 72 | + |
| 73 | +```python |
| 74 | +class Solution: |
| 75 | + def permute(self, n: int) -> List[List[int]]: |
| 76 | + def dfs(i: int) -> None: |
| 77 | + if i >= n: |
| 78 | + ans.append(t[:]) |
| 79 | + return |
| 80 | + for j in range(1, n + 1): |
| 81 | + if not vis[j] and (i == 0 or t[-1] % 2 != j % 2): |
| 82 | + t.append(j) |
| 83 | + vis[j] = True |
| 84 | + dfs(i + 1) |
| 85 | + vis[j] = False |
| 86 | + t.pop() |
| 87 | + |
| 88 | + ans = [] |
| 89 | + t = [] |
| 90 | + vis = [False] * (n + 1) |
| 91 | + dfs(0) |
| 92 | + return ans |
| 93 | +``` |
| 94 | + |
| 95 | +#### Java |
| 96 | + |
| 97 | +```java |
| 98 | +class Solution { |
| 99 | + private List<int[]> ans = new ArrayList<>(); |
| 100 | + private boolean[] vis; |
| 101 | + private int[] t; |
| 102 | + private int n; |
| 103 | + |
| 104 | + public int[][] permute(int n) { |
| 105 | + this.n = n; |
| 106 | + t = new int[n]; |
| 107 | + vis = new boolean[n + 1]; |
| 108 | + dfs(0); |
| 109 | + return ans.toArray(new int[0][]); |
| 110 | + } |
| 111 | + |
| 112 | + private void dfs(int i) { |
| 113 | + if (i >= n) { |
| 114 | + ans.add(t.clone()); |
| 115 | + return; |
| 116 | + } |
| 117 | + for (int j = 1; j <= n; ++j) { |
| 118 | + if (!vis[j] && (i == 0 || t[i - 1] % 2 != j % 2)) { |
| 119 | + vis[j] = true; |
| 120 | + t[i] = j; |
| 121 | + dfs(i + 1); |
| 122 | + vis[j] = false; |
| 123 | + } |
| 124 | + } |
| 125 | + } |
| 126 | +} |
| 127 | +``` |
| 128 | + |
| 129 | +#### C++ |
| 130 | + |
| 131 | +```cpp |
| 132 | +class Solution { |
| 133 | +public: |
| 134 | + vector<vector<int>> permute(int n) { |
| 135 | + vector<vector<int>> ans; |
| 136 | + vector<bool> vis(n); |
| 137 | + vector<int> t; |
| 138 | + auto dfs = [&](this auto&& dfs, int i) -> void { |
| 139 | + if (i >= n) { |
| 140 | + ans.push_back(t); |
| 141 | + return; |
| 142 | + } |
| 143 | + for (int j = 1; j <= n; ++j) { |
| 144 | + if (!vis[j] && (i == 0 || t[i - 1] % 2 != j % 2)) { |
| 145 | + vis[j] = true; |
| 146 | + t.push_back(j); |
| 147 | + dfs(i + 1); |
| 148 | + t.pop_back(); |
| 149 | + vis[j] = false; |
| 150 | + } |
| 151 | + } |
| 152 | + }; |
| 153 | + dfs(0); |
| 154 | + return ans; |
| 155 | + } |
| 156 | +}; |
| 157 | +``` |
| 158 | + |
| 159 | +#### Go |
| 160 | + |
| 161 | +```go |
| 162 | +func permute(n int) (ans [][]int) { |
| 163 | + vis := make([]bool, n+1) |
| 164 | + t := make([]int, n) |
| 165 | + var dfs func(i int) |
| 166 | + dfs = func(i int) { |
| 167 | + if i >= n { |
| 168 | + ans = append(ans, slices.Clone(t)) |
| 169 | + return |
| 170 | + } |
| 171 | + for j := 1; j <= n; j++ { |
| 172 | + if !vis[j] && (i == 0 || t[i-1]%2 != j%2) { |
| 173 | + vis[j] = true |
| 174 | + t[i] = j |
| 175 | + dfs(i + 1) |
| 176 | + vis[j] = false |
| 177 | + } |
| 178 | + } |
| 179 | + } |
| 180 | + dfs(0) |
| 181 | + return |
| 182 | +} |
| 183 | +``` |
| 184 | + |
| 185 | +#### TypeScript |
| 186 | + |
| 187 | +```ts |
| 188 | +function permute(n: number): number[][] { |
| 189 | + const ans: number[][] = []; |
| 190 | + const vis: boolean[] = Array(n).fill(false); |
| 191 | + const t: number[] = Array(n).fill(0); |
| 192 | + const dfs = (i: number) => { |
| 193 | + if (i >= n) { |
| 194 | + ans.push([...t]); |
| 195 | + return; |
| 196 | + } |
| 197 | + for (let j = 1; j <= n; ++j) { |
| 198 | + if (!vis[j] && (i === 0 || t[i - 1] % 2 !== j % 2)) { |
| 199 | + vis[j] = true; |
| 200 | + t[i] = j; |
| 201 | + dfs(i + 1); |
| 202 | + vis[j] = false; |
| 203 | + } |
| 204 | + } |
| 205 | + }; |
| 206 | + dfs(0); |
| 207 | + return ans; |
| 208 | +} |
| 209 | +``` |
| 210 | + |
| 211 | +<!-- tabs:end --> |
| 212 | + |
| 213 | +<!-- solution:end --> |
| 214 | + |
| 215 | +<!-- problem:end --> |
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