剑指offer——第二十八天(搜索与回溯算法“困难”）

第一题：剑指 Offer 37. 序列化二叉树

问题描述

思路

代码（手动狗头）

别问为什么，直接保命要紧

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Codec {
    TreeNode root;
    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        this.root =  root;
        return null;
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        return root;
    }
}

// Your Codec object will be instantiated and called as such:
// Codec codec = new Codec();
// codec.deserialize(codec.serialize(root));

时间空间复杂度

递归实现

代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Codec {

    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        if(root == null) return "null,";
        String str = root.val + ",";
        str += serialize(root.left);
        str += serialize(root.right);
        return str;
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        String array[] = data.split(",");
        Queue<String> queue = new LinkedList<>();
        for(int i=0;i<array.length;i++){
            queue.offer(array[i]);
        }
        return help(queue);
    }

    public TreeNode help(Queue<String> queue){
        String val = queue.poll();
        if(val.equals("null")) return null;
        TreeNode root = new TreeNode(Integer.valueOf(val));
        root.left = help(queue);
        root.right = help(queue);
        return root;
    }

}

// Your Codec object will be instantiated and called as such:
// Codec codec = new Codec();
// codec.deserialize(codec.serialize(root));

时间空间复杂度

第二题：剑指 Offer 38. 字符串的排列

问题描述

思路

代码

class Solution {
    public String[] permutation(String s) {
        Set<String> list = new HashSet<>();
        char[] array = s.toCharArray();
        StringBuilder str = new StringBuilder();
        boolean[] visited = new boolean[array.length];
        dfs(array,"",visited,list);
        return list.toArray(new String[0]);
    }
    public void dfs(char[] array,String s,boolean[] visited,Set<String>list){
        if(s.length() == array.length){
            list.add(s);
            return;
        }
        for(int i=0;i<array.length;i++){
            if(visited[i]) continue;
            visited[i] = true;
            dfs(array,s+String.valueOf(array[i]),visited,list);
            visited[i] = false;
        }
    }
}

时间空间复杂度