剑指 Offer 37. 序列化二叉树

请实现两个函数，分别用来序列化和反序列化二叉树。

• 层序遍历

/**
 * 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 "[]";
        }
       
       //1.得到层序遍历列表
        List<String> nodeList = new ArrayList<>();
        Deque<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while(!queue.isEmpty()){
            TreeNode node = queue.poll();
            if(node != null){
                nodeList.add(String.valueOf(node.val));
                queue.offer(node.left);
                queue.offer(node.right);
            }else{
                nodeList.add("null");
            }
        }

        //2.去掉最末尾的null
        int right = nodeList.size() - 1;
        while(right > 0){
            if(!"null".equals(nodeList.get(right))){
                break;
            }
            right--;
        }
        //3.拼接结果字符串
        StringBuilder sb = new StringBuilder();
        sb.append("[");
        for(int i = 0; i <= right; i++){
            if(i != 0){
                sb.append(",");
            }
            sb.append(nodeList.get(i));
        }
        sb.append("]");
        return sb.toString();
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        String[] nodeValues = data.substring(1, data.length() - 1).split(",");
        int len = nodeValues.length;
        //System.out.println(len);
        if(len == 0 || data.equals("[]")){
            return null;
        }
        
        TreeNode root = new TreeNode(Integer.valueOf(nodeValues[0]));
        Queue<TreeNode> nodes = new LinkedList<>();
        //根节点入队
        nodes.offer(root);
        int i = 1;
        
        while(!nodes.isEmpty() && i < len){
            TreeNode temp = nodes.poll();
            if(!nodeValues[i].equals("null")){
                temp.left = new TreeNode(Integer.valueOf(nodeValues[i]));
                nodes.offer(temp.left);
            }
            i++;
            if(i < len && !nodeValues[i].equals("null")){
                temp.right = new TreeNode(Integer.valueOf(nodeValues[i]));
                nodes.offer(temp.right);
            }
            i++;
        }
        return root; 
    }
}

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

剑指 Offer 38. 字符串的排列

输入一个字符串，打印出该字符串中字符的所有排列。
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。

• 典型的全排列问题，采用回溯法；哈希表去重

class Solution {
    HashSet<String> res = new HashSet<>();
    char[] chars;
    int[] visited;
    public String[] permutation(String s) {
        chars = s.toCharArray();
        Arrays.sort(chars);
        visited = new int[chars.length];

        List<Character> track = new ArrayList<>();
        dfs(track);
        
        String[] resStr = new String[res.size()];
        int i = 0;
        for(String str : res){
            resStr[i++] = str;
        }
        return resStr;
    }

    public void dfs(List<Character> track) {
        if(track.size() == chars.length){
            String s = list2Str(track);
            if(!res.contains(s)){
                res.add(s);
            }
            return;
        }

        for(int i = 0; i < chars.length; i++){
            if(visited[i] == 1){
                continue;
            }

            track.add(chars[i]);
            visited[i] = 1;

            dfs(track);

            track.remove(track.size() - 1);
            visited[i] = 0;
        }
    }

    public String list2Str(List<Character> track){
        StringBuilder sb = new StringBuilder();
        for(Character c : track){
            sb.append(c);
        }
        return sb.toString();
    }
}

剑指 Offer 39. 数组中出现次数超过一半的数字

数组中有一个数字出现的次数超过数组长度的一半，请找出这个数字。
你可以假设数组是非空的，并且给定的数组总是存在多数元素。

• 排序后返回最中间的元素

class Solution {
    public int majorityElement(int[] nums) {
        int len = nums.length;
        Arrays.sort(nums);
        return nums[len/2];
    }
}

剑指 Offer 40. 最小的k个数

输入整数数组 arr ，找出其中最小的 k 个数。例如，输入4、5、1、6、2、7、3、8这8个数字，则最小的4个数字是1、2、3、4。

• 调用Array.sort

class Solution {
    public int[] getLeastNumbers(int[] arr, int k) {
        int[] res = new int[k];
        Arrays.sort(arr);
        for(int i = 0; i < k; i++){
            res[i] = arr[i];
        }
        return res;
    }
}

• 基于快排

class Solution {
    public int[] getLeastNumbers(int[] arr, int k) {
        quickSort(arr, 0 , arr.length - 1, k);
        int[] res = new int[k];
        for(int i = 0; i < k; i++){
            res[i] = arr[i];
        }
        return res;
        
    }

    public void quickSort(int[] arr, int left , int right,  int k){
        if(left >= right){
            return;
        }

        int index = partition(arr, left, right);
        if(left >= k){
            return;
        }
        quickSort(arr, left, index - 1, k);
        quickSort(arr, index + 1, right, k);
    }

    public int partition(int[] arr, int left , int right){
        int temp = arr[left];
        while(left < right){
            while(left < right && arr[right] > temp){
                right--;
            }
            arr[left] = arr[right];
            while(left < right && arr[left] <= temp){
                left++;
            }
            arr[right] = arr[left];
        }
        arr[left] = temp;
        return left;
    }
}