字符串相乘

43. 字符串相乘

难度中等740

给定两个以字符串形式表示的非负整数 num1 和 num2，返回 num1 和 num2 的乘积，它们的乘积也表示为字符串形式。

示例 1:

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输入: num1 = "2", num2 = "3"
输出: "6"

示例 2:

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输入: num1 = "123", num2 = "456"
输出: "56088"

说明：

  1. num1 和 num2 的长度小于110。
  2. num1 和 num2 只包含数字 0-9。
  3. num1 和 num2 均不以零开头，除非是数字 0 本身。
  4. 不能使用任何标准库的大数类型（比如 BigInteger）或直接将输入转换为整数来处理。

通过次数174,069

提交次数387,591

请问您在哪类招聘中遇到此题？

从后往前做乘法

  • 第i+j位是进位

  • 第i+j+1位是要计算的位

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class Solution {
    public String multiply(String num1, String num2) {
        int n1 = num1.length();
        int n2 = num2.length();
        int [] arr = new int[n1+n2];
        for(int i=n1-1;i>=0;i--){
            for(int j=n2-1;j>=0;j--){
                int num = (num1.charAt(i)-'0')*(num2.charAt(j)-'0');
                // 先加上当前的数字
                arr[i+j+1]+=num;
                // 处理进位
                arr[i+j] += arr[i+j+1]/10;
                // 进位后处理
                arr[i+j+1] = arr[i+j+1]%10;
            }
        }

        StringBuilder bd = new StringBuilder();
        // 处理前缀0
        int index =0;
        while(index<n1+n2&&arr[index]==0)index++;
        for(int i=index;i<n1+n2;i++)bd.append(arr[i]);
        // 处理结果是0
        if(bd.length()==0) bd.append(0);
        return bd.toString();
    }
}

展开全文 >>

链表内指定区间反转

描述

将一个节点数为 size 链表 m 位置到 n 位置之间的区间反转，要求时间复杂度 O(n)O(n)，空间复杂度 O(1)O(1)。
例如：
给出的链表为 1\to 2 \to 3 \to 4 \to 5 \to NULL1→2→3→4→5→NUL**L, m=2,n=4m=2,n=4,
返回 1\to 4\to 3\to 2\to 5\to NULL1→4→3→2→5→NUL**L.

数据范围： 链表长度 0 < size \le 10000<s*ize≤1000，0 < m \le n \le size0<m≤n≤size，链表中每个节点的值满足 |val| \le 1000∣va*l∣≤1000

要求：时间复杂度 O(n)O(n) ，空间复杂度 O(n)O(n)

进阶：时间复杂度 O(n)O(n)，空间复杂度 O(1)O(1)

示例1

输入：

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{1,2,3,4,5},2,4

返回值：

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{1,4,3,2,5}

示例2

输入：

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{5},1,1

返回值：

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{5}

模拟

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import java.util.*;

public class Solution {

    public ListNode reverseBetween (ListNode head, int m, int n) {
        ListNode cur = head;
        ListNode dummyNode = new ListNode(0);
        dummyNode.next = head;
        ListNode pre = dummyNode;
        // m节点
        for(int i=1;i<m;i++){
            pre = cur;
            cur = cur.next;
        }
        
        // n之后的第一个节点
        ListNode next = null;
        for(int i=0;i<=n-m;i++){
            cur = cur.next;
            next = cur;
        }
        
        ListNode node = pre;
        cur = pre.next;
        
        // 反转后的尾节点就是当前的头节点
        ListNode tail = cur;
        pre = null;
        
        while(cur!=next){
            ListNode nex = cur.next;
            cur.next = pre;
            pre = cur;
            cur = nex;
        }
        
        node.next = pre;
        tail.next = next;
        
        return dummyNode.next;
    }
}

展开全文 >>

二叉搜索树中第k小的元素

230. 二叉搜索树中第K小的元素

难度中等494

给定一个二叉搜索树的根节点 root ，和一个整数 k ，请你设计一个算法查找其中第 k 个最小元素（从 1 开始计数）。

示例 1：

img

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输入：root = [3,1,4,null,2], k = 1
输出：1

示例 2：

img

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输入：root = [5,3,6,2,4,null,null,1], k = 3
输出：3

提示：

  • 树中的节点数为 n 。
  • 1 <= k <= n <= 104
  • 0 <= Node.val <= 104

进阶：如果二叉搜索树经常被修改（插入/删除操作）并且你需要频繁地查找第 k 小的值，你将如何优化算法？

通过次数150,512

提交次数201,309

中序遍历

二叉搜索树的中序遍历是从小到大的

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    
    public int kthSmallest(TreeNode root, int k) {
        dfs(root,k);
        return ans;
    }
    int cnt = 0;
    int ans = 0;
    public void dfs(TreeNode root,int k){
        if(root==null)return;
        dfs(root.left,k);
        cnt++;
        if(cnt==k){
            ans = root.val;
            return;
        }
        dfs(root.right,k);
    }
}

展开全文 >>

外观数列

38. 外观数列

难度中等813

给定一个正整数 n ，输出外观数列的第 n 项。

「外观数列」是一个整数序列，从数字 1 开始，序列中的每一项都是对前一项的描述。

你可以将其视作是由递归公式定义的数字字符串序列：

  • countAndSay(1) = "1"
  • countAndSay(n) 是对 countAndSay(n-1) 的描述，然后转换成另一个数字字符串。

前五项如下：

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1.     1
2.     11
3.     21
4.     1211
5.     111221
第一项是数字 1 
描述前一项，这个数是 1 即 “ 一 个 1 ”，记作 "11"
描述前一项，这个数是 11 即 “ 二 个 1 ” ，记作 "21"
描述前一项，这个数是 21 即 “ 一 个 2 + 一 个 1 ” ，记作 "1211"
描述前一项，这个数是 1211 即 “ 一 个 1 + 一 个 2 + 二 个 1 ” ，记作 "111221"

要 描述 一个数字字符串，首先要将字符串分割为 最小 数量的组，每个组都由连续的最多 相同字符 组成。然后对于每个组，先描述字符的数量，然后描述字符，形成一个描述组。要将描述转换为数字字符串，先将每组中的字符数量用数字替换，再将所有描述组连接起来。

例如，数字字符串 "3322251" 的描述如下图：

img

示例 1：

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输入：n = 1
输出："1"
解释：这是一个基本样例。

示例 2：

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输入：n = 4
输出："1211"
解释：
countAndSay(1) = "1"
countAndSay(2) = 读 "1" = 一 个 1 = "11"
countAndSay(3) = 读 "11" = 二 个 1 = "21"
countAndSay(4) = 读 "21" = 一 个 2 + 一 个 1 = "12" + "11" = "1211"

提示：

  • 1 <= n <= 30

通过次数239,119

提交次数402,335

模拟

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class Solution {
    public String countAndSay(int n) {
        StringBuilder bd = new StringBuilder();
        bd.append(1);

        for(int i=2;i<=n;i++){
            StringBuilder bd1 = new StringBuilder();
            int last = bd.charAt(0);
            int cnt = 1;
            
            for(int j=1;j<bd.length();j++){
                if(bd.charAt(j)==last){
                    cnt++;
                }else{
                    bd1.append(cnt);
                    bd1.append(last-'0');
                    last = bd.charAt(j);
                    cnt = 1;   
                }
            }
            bd1.append(cnt);
            bd1.append(last-'0');
            

            bd = bd1;
        }
        return bd.toString();
    }
}

打表法

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class Solution {
    public String countAndSay(int n) {
        String[] arr = {
            "","1","11","21","1211","111221","312211","13112221","1113213211","31131211131221","13211311123113112211","11131221133112132113212221","3113112221232112111312211312113211","1321132132111213122112311311222113111221131221","11131221131211131231121113112221121321132132211331222113112211","311311222113111231131112132112311321322112111312211312111322212311322113212221","132113213221133112132113311211131221121321131211132221123113112221131112311332111213211322211312113211","11131221131211132221232112111312212321123113112221121113122113111231133221121321132132211331121321231231121113122113322113111221131221","31131122211311123113321112131221123113112211121312211213211321322112311311222113311213212322211211131221131211132221232112111312111213111213211231131122212322211331222113112211","1321132132211331121321231231121113112221121321132122311211131122211211131221131211132221121321132132212321121113121112133221123113112221131112311332111213122112311311123112111331121113122112132113213211121332212311322113212221","11131221131211132221232112111312111213111213211231132132211211131221131211221321123113213221123113112221131112311332211211131221131211132211121312211231131112311211232221121321132132211331121321231231121113112221121321133112132112312321123113112221121113122113121113123112112322111213211322211312113211","311311222113111231133211121312211231131112311211133112111312211213211312111322211231131122211311122122111312211213211312111322211213211321322113311213212322211231131122211311123113223112111311222112132113311213211221121332211211131221131211132221232112111312111213111213211231132132211211131221232112111312211213111213122112132113213221123113112221131112311311121321122112132231121113122113322113111221131221","132113213221133112132123123112111311222112132113311213211231232112311311222112111312211311123113322112132113213221133122112231131122211211131221131112311332211211131221131211132221232112111312111213322112132113213221133112132113221321123113213221121113122123211211131221222112112322211231131122211311123113321112131221123113111231121113311211131221121321131211132221123113112211121312211231131122211211133112111311222112111312211312111322211213211321322113311213211331121113122122211211132213211231131122212322211331222113112211","111312211312111322212321121113121112131112132112311321322112111312212321121113122112131112131221121321132132211231131122211331121321232221121113122113121113222123112221221321132132211231131122211331121321232221123113112221131112311332111213122112311311123112112322211211131221131211132221232112111312211322111312211213211312111322211231131122111213122112311311221132211221121332211213211321322113311213212312311211131122211213211331121321123123211231131122211211131221131112311332211213211321223112111311222112132113213221123123211231132132211231131122211311123113322112111312211312111322212321121113122123211231131122113221123113221113122112132113213211121332212311322113212221","3113112221131112311332111213122112311311123112111331121113122112132113121113222112311311221112131221123113112221121113311211131122211211131221131211132221121321132132212321121113121112133221123113112221131112311332111213213211221113122113121113222112132113213221232112111312111213322112132113213221133112132123123112111311222112132113311213211221121332211231131122211311123113321112131221123113112221132231131122211211131221131112311332211213211321223112111311222112132113212221132221222112112322211211131221131211132221232112111312111213111213211231132132211211131221232112111312211213111213122112132113213221123113112221133112132123222112111312211312112213211231132132211211131221131211132221121311121312211213211312111322211213211321322113311213212322211231131122211311123113321112131221123113112211121312211213211321222113222112132113223113112221121113122113121113123112112322111213211322211312113211","1321132132211331121321231231121113112221121321133112132112312321123113112221121113122113111231133221121321132122311211131122211213211321322112312321123113213221123113112221131112311332211211131221131211132211121312211231131112311211232221121321132132211331121321231231121113121113122122311311222113111231133221121113122113121113221112131221123113111231121123222112111312211312111322212321121113121112131112132112311321322112111312212321121113122122211211232221121321132132211331121321231231121113112221121321132132211322132113213221123113112221133112132123222112111312211312112213211231132132211211131221131211322113321132211221121332211231131122211311123113321112131221123113111231121113311211131221121321131211132221123113112211121312211231131122211211133112111311222112111312211312111322211213211321322123211211131211121332211231131122211311122122111312211213211312111322211231131122211311123113322112111331121113112221121113122113111231133221121113122113121113222123211211131211121332211213211321322113311213212312311211131122211213211321223112111311222112111312211312113221133221121113122113221321132132211231131122211311123113111213211221121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        };
        return arr[n];
    }
}

作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/count-and-say/solution/wai-guan-shu-lie-by-leetcode-solution-9rt8/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

展开全文 >>

前n个数组二进制中1的个数

剑指 Offer II 003. 前 n 个数字二进制中 1 的个数

难度简单13

给定一个非负整数 n ，请计算 0 到 n 之间的每个数字的二进制表示中 1 的个数，并输出一个数组。

示例 1:

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输入: n = 2
输出: [0,1,1]
解释: 
0 --> 0
1 --> 1
2 --> 10

示例 2:

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输入: n = 5
输出: [0,1,1,2,1,2]
解释:
0 --> 0
1 --> 1
2 --> 10
3 --> 11
4 --> 100
5 --> 101

说明 :

  • 0 <= n <= 105

进阶:

  • 给出时间复杂度为 O(n*sizeof(integer)) 的解答非常容易。但你可以在线性时间 O(n) 内用一趟扫描做到吗？

  • 要求算法的空间复杂度为 O(n) 。

  • 你能进一步完善解法吗？要求在C++或任何其他语言中不使用任何内置函数（如 C++ 中的 __builtin_popcount ）来执行此操作。

    java bitcount解法

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class Solution {
    public int[] countBits(int n) {
        
        int []ans = new int[n+1];
        for(int i=0;i<=n;i++){
            ans[i] = bitCount(i);
        }
        return ans;
    }


    public static int bitCount(int i) {
        // HD, Figure 5-2
        i = i - ((i >>> 1) & 0x55555555);
        i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
        i = (i + (i >>> 4)) & 0x0f0f0f0f;
        i = i + (i >>> 8);
        i = i + (i >>> 16);
        return i & 0x3f;
    }

}

动态规划 +位运算

对于所有的数字，只有奇数和偶数两种：

奇数：二进制表示中，奇数一定比前面那个偶数多一个 1，因为多的就是最低位的 1。
偶数：二进制表示中，偶数中 1 的个数一定和除以 2 之后的那个数一样多。因为最低位是 0，除以 2 就是右移一位，也就是把那个 0 抹掉而已，所以 1 的个数是不变的。
所以我们可以得到如下的状态转移方程：

$dp[i] = dp[i-1]$，当i为奇数
$dp[i] = dp[i/2]$，当i为偶数
上面的方程还可进一步合并为：
$dp[i] = dp[i/2] + i % 2$

通过位运算进一步优化：

$i / 2 $可以通过$ i >> 1 $得到；
$i % 2 $可以通过$ i & 1 $得到；

作者：Terry2020
链接：https://leetcode-cn.com/problems/w3tCBm/solution/rang-ni-miao-dong-de-shuang-bai-ti-jie-b-84hh/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

展开全文 >>

允许重复选择元素的组合

剑指 Offer II 081. 允许重复选择元素的组合

难度中等3

给定一个无重复元素的正整数数组 candidates 和一个正整数 target ，找出 candidates 中所有可以使数字和为目标数 target 的唯一组合。

candidates 中的数字可以无限制重复被选取。如果至少一个所选数字数量不同，则两种组合是唯一的。

对于给定的输入，保证和为 target 的唯一组合数少于 150 个。

示例 1：

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输入: candidates = [2,3,6,7], target = 7
输出: [[7],[2,2,3]]

示例 2：

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输入: candidates = [2,3,5], target = 8
输出: [[2,2,2,2],[2,3,3],[3,5]]

示例 3：

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输入: candidates = [2], target = 1
输出: []

示例 4：

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输入: candidates = [1], target = 1
输出: [[1]]

示例 5：

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输入: candidates = [1], target = 2
输出: [[1,1]]

提示：

  • 1 <= candidates.length <= 30
  • 1 <= candidates[i] <= 200
  • candidate 中的每个元素都是独一无二的。
  • 1 <= target <= 500

注意：本题与主站 39 题相同： https://leetcode-cn.com/problems/combination-sum/

回溯

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class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        backtrack(candidates,0,target,new ArrayList<>());
        return ans;
    }

    public List<List<Integer>> ans = new ArrayList<>();
    public void backtrack(int []candidates,int sum,int target,List<Integer> list){
        if(sum > target)return;
        if(sum==target){
            List<Integer> temp =  new ArrayList(list);
            Collections.sort(temp);
            if(!ans.contains(temp)){
                ans.add(temp);
            }
            return;
        }
        for(int i=0;i<candidates.length;i++){   
            list.add(candidates[i]);
            sum+=candidates[i];
            backtrack(candidates,sum,target,list);
            sum-=candidates[i];
            list.remove(list.size()-1);
        }
    }
}

回溯剪枝

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class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> ans = new ArrayList<List<Integer>>();
        List<Integer> combine = new ArrayList<Integer>();
        dfs(candidates, target, ans, combine, 0);
        return ans;
    }

    public void dfs(int[] candidates, int target, List<List<Integer>> ans, List<Integer> combine, int idx) {
        if (idx == candidates.length) {
            return;
        }
        if (target == 0) {
            ans.add(new ArrayList<Integer>(combine));
            return;
        }
        // 直接跳过
        dfs(candidates, target, ans, combine, idx + 1);
        // 选择当前数
        if (target - candidates[idx] >= 0) {
            combine.add(candidates[idx]);
            dfs(candidates, target - candidates[idx], ans, combine, idx);
            combine.remove(combine.size() - 1);
        }
    }
}

作者：vanilla111
链接：https://leetcode-cn.com/problems/Ygoe9J/solution/java-dfsjian-zhi-tao-lu-by-vanilla111-8lpp/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

展开全文 >>

山峰数组的顶部

剑指 Offer II 069. 山峰数组的顶部

难度简单27

符合下列属性的数组 arr 称为 山峰数组（山脉数组） ：

  • arr.length >= 3

  • 存在（0 < i < arr.length - 1）使得：

    • arr[0] < arr[1] < ... arr[i-1] < arr[i]
    • arr[i] > arr[i+1] > ... > arr[arr.length - 1]

给定由整数组成的山峰数组 arr ，返回任何满足 arr[0] < arr[1] < ... arr[i - 1] < arr[i] > arr[i + 1] > ... > arr[arr.length - 1] 的下标 i ，即山峰顶部。

示例 1：

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输入：arr = [0,1,0]
输出：1

示例 2：

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输入：arr = [1,3,5,4,2]
输出：2

示例 3：

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输入：arr = [0,10,5,2]
输出：1

示例 4：

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输入：arr = [3,4,5,1]
输出：2

示例 5：

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输入：arr = [24,69,100,99,79,78,67,36,26,19]
输出：2

提示：

  • 3 <= arr.length <= 104
  • 0 <= arr[i] <= 106
  • 题目数据保证 arr 是一个山脉数组

进阶：很容易想到时间复杂度 O(n) 的解决方案，你可以设计一个 O(log(n)) 的解决方案吗？

注意：本题与主站 852 题相同：https://leetcode-cn.com/problems/peak-index-in-a-mountain-array/

二分查找

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class Solution {
    public int peakIndexInMountainArray(int[] arr) {
        int n = arr.length;
        int l = 0, r = n-1;
        // 处理边界条件
        if(arr[0]>arr[1]) return 0;
        if(arr[n-1]>arr[n-2]) return n - 1;
        // 进行二分查找
        while(l<r){
            int m = l+(r-l)/2;
            if(m>0 && m<n-1 && arr[m-1]<arr[m] && arr[m]>arr[m+1]){
                return m;
            }else if(m<n-1 && arr[m]<arr[m+1]){
                l = m+1;
            }else if(m<n-1 && arr[m]>arr[m+1]){
                r = m-1;
            }
        }
        return l;
    }
}

官方题解二分查找

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class Solution {
    public int peakIndexInMountainArray(int[] arr) {
        int n = arr.length;
        int left = 1, right = n - 2, ans = 0;
        while (left <= right) {
            int mid = (left + right) / 2;
            if (arr[mid] > arr[mid + 1]) {
                ans = mid;
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        return ans;
    }
}

作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/B1IidL/solution/shan-feng-shu-zu-de-ding-bu-by-leetcode-9j8lk/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

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fizz_Buzz

412. Fizz Buzz

难度简单126

写一个程序，输出从 1 到 n 数字的字符串表示。

\1. 如果 n 是3的倍数，输出“Fizz”；

\2. 如果 n 是5的倍数，输出“Buzz”；

3.如果 n 同时是3和5的倍数，输出 “FizzBuzz”。

示例：

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n = 15,

返回:
[
    "1",
    "2",
    "Fizz",
    "4",
    "Buzz",
    "Fizz",
    "7",
    "8",
    "Fizz",
    "Buzz",
    "11",
    "Fizz",
    "13",
    "14",
    "FizzBuzz"
]

通过次数84,968

提交次数122,924

模拟

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class Solution {
    public List<String> fizzBuzz(int n) {
        List<String> list = new ArrayList<>();
        for(int i=1;i<=n;i++){
            if(i%3==0&&i%5==0){
                list.add("FizzBuzz");
            }else if(i%3==0){
                list.add("Fizz");
            }else if(i%5==0){
                list.add("Buzz");
            }else{
                list.add(String.valueOf(i));
            }
        }
        return list;
    }
}

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树洞202110

2021.10.11

有一个“刺猬效应”是这样说的：

人与人之间，只有保持适当的距离，才能最大限度地感受对方的美好。

亲密有间，疏而不远，才是关系长久稳定的良方。

愿你我都能，做恰到好处的事，当行事有度的人。

展开全文 >>

tag:

    缺失模块。
    1、请确保node版本大于6.2
    2、在博客根目录（注意不是yilia根目录）执行以下命令：
    npm i hexo-generator-json-content --save

    3、在根目录_config.yml里添加配置： 

      jsonContent:
    meta: false
    pages: false
    posts:
      title: true
      date: true
      path: true
      text: false
      raw: false
      content: false
      slug: false
      updated: false
      comments: false
      link: false
      permalink: false
      excerpt: false
      categories: false
      tags: true

我是一只<br>又水又笨又菜的<br>菜狗子<br>希望有一天被领走