剑指 Offer 38:字符串的排列
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src/main/java/leetcode/editor/cn/ZiFuChuanDePaiLieLcof.java
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src/main/java/leetcode/editor/cn/ZiFuChuanDePaiLieLcof.java
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//输入一个字符串,打印出该字符串中字符的所有排列。
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//
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//
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//
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// 你可以以任意顺序返回这个字符串数组,但里面不能有重复元素。
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//
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//
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//
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// 示例:
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//
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// 输入:s = "abc"
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//输出:["abc","acb","bac","bca","cab","cba"]
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//
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//
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//
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//
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// 限制:
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//
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// 1 <= s 的长度 <= 8
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// Related Topics 回溯算法
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// 👍 327 👎 0
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package leetcode.editor.cn;
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import java.util.HashSet;
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import java.util.LinkedList;
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import java.util.List;
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//剑指 Offer 38:字符串的排列
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public class ZiFuChuanDePaiLieLcof {
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public static void main(String[] args) {
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//测试代码
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Solution solution = new ZiFuChuanDePaiLieLcof().new Solution();
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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class Solution {
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List<String> res = new LinkedList<>();
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char[] c;
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public String[] permutation(String s) {
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c = s.toCharArray();
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dfs(0);
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return res.toArray(new String[res.size()]);
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}
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void dfs(int x) {
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if (x == c.length - 1) {
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res.add(String.valueOf(c)); // 添加排列方案
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return;
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}
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HashSet<Character> set = new HashSet<>();
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for (int i = x; i < c.length; i++) {
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if (set.contains(c[i])) {
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continue; // 重复,因此剪枝
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}
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set.add(c[i]);
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swap(i, x); // 交换,将 c[i] 固定在第 x 位
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dfs(x + 1); // 开启固定第 x + 1 位字符
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swap(i, x); // 恢复交换
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}
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}
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void swap(int a, int b) {
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char tmp = c[a];
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c[a] = c[b];
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c[b] = tmp;
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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}
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src/main/java/leetcode/editor/cn/ZiFuChuanDePaiLieLcof.md
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src/main/java/leetcode/editor/cn/ZiFuChuanDePaiLieLcof.md
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<p>输入一个字符串,打印出该字符串中字符的所有排列。</p>
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<p> </p>
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<p>你可以以任意顺序返回这个字符串数组,但里面不能有重复元素。</p>
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<p> </p>
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<p><strong>示例:</strong></p>
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<pre><strong>输入:</strong>s = "abc"
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<strong>输出:[</strong>"abc","acb","bac","bca","cab","cba"<strong>]</strong>
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</pre>
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<p> </p>
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<p><strong>限制:</strong></p>
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<p><code>1 <= s 的长度 <= 8</code></p>
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<div><div>Related Topics</div><div><li>回溯算法</li></div></div>\n<div><li>👍 327</li><li>👎 0</li></div>
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