Awesome solutions for algorithm questions

你就会发现只要涉及递归的问题,都是 树的问题。

Different Trees

  • Full Binary Tree : For every node, it can either has no children or two children
  • Complete Binary Tree: Leaf nodes are aligned leftwards (it must be left child if there is only one child)
  • Perfect Binary Tree: every node has two children

但是必须说明的是,不管怎么优化,都符合回溯框架,而且时间复杂度都不 可能低于 O(N!),因为穷举整棵决策树是无法避免的。这也是回溯算法的一 个特点,不像动态规划存在重叠子问题可以优化,回溯算法就是纯暴力穷 举,复杂度一般都很高。

vector<vector<string>> res;
/* 输入棋盘边⻓ n,返回所有合法的放置 */ vector<vector<string>> solveNQueens(int n) {
// '.' 表示空,'Q' 表示皇后,初始化空棋盘。 vector<string> board(n, string(n, '.')); backtrack(board, 0);
return res;
}
// 路径:board 中小于 row 的那些行都已经成功放置了皇后 // 选择列表:第 row 行的所有列都是放置皇后的选择
// 结束条件:row 超过 board 的最后一行
void backtrack(vector<string>& board, int row) {
// 触发结束条件
if (row == board.size()) {
res.push_back(board);
return; }
int n = board[row].size();
for (int col = 0; col < n; col++) {
// 排除不合法选择
if (!isValid(board, row, col))
continue; // 做选择
board[row][col] = 'Q';
// 进入下一行决策 backtrack(board, row + 1); // 撤销选择
board[row][col] = '.';

}
}
/* 是否可以在 board[row][col] 放置皇后? */
bool isValid(vector<string>& board, int row, int col) {
int n = board.size();
// 检查列是否有皇后互相冲突
for (int i = 0; i < n; i++) {
    if (board[i][col] == 'Q')
        return false;
}
// 检查右上方是否有皇后互相冲突 for (int i = row - 1, j = col i >= 0 && j < n; i--,
+ 1; j++) {
    if (board[i][j] == 'Q')
        return false;
}
// 检查左上方是否有皇后互相冲突 for (int i = row - 1, j = col
- 1;
i >= 0 && j >= 0; i--, j--) {
        if (board[i][j] == 'Q')
            return false;
}
    return true;
}

Suduko

有的时候,我们并不想得到所有合法的答案,只想要一个答案,怎么办呢? 比如解数独的算法,找所有解法复杂度太高,只要找到一种解法就可以。 其实特别简单,只要稍微修改一下回溯算法的代码即可:

Backtrack summary

写 backtrack 函数时,需要维护走过的「路径」和当前可以做的「选择列 表」,当触发「结束条件」时,将「路径」记入结果集。 其实想想看,回溯算法和动态规划是不是有点像呢?我们在动态规划系列文 章中多次强调,动态规划的三个需要明确的点就是「状态」「选择」和 「base case」,是不是就对应着走过的「路径」,当前的「选择列表」和 「结束条件」?

Binary Search

分析二分查找的一个技巧是:不要出现 else,而是把所有情况用 else if 写清 楚,这样可以清楚地展现所有细节。本文都会使用 else if,旨在讲清楚,读 者理解后可自行简化。

寻找左侧边界的二分搜索

以下是最常⻅的代码形式,其中的标记是需要注意的细节:

int left_bound(int[] nums, int target)
{
  if (nums.length == 0) return -1;
  int left = 0;
  int right = nums.length; // 注意
  while (left < right) { // 注意
    int mid = (left + right) / 2;
    if (nums[mid] == target) {
            right = mid;
    } else if (nums[mid] < target) {
            left = mid + 1;
    } else if (nums[mid] > target) {
      right = mid; // 注意 }
    }
  }
    return left;
}

Left left_bound

int left_bound(int[] nums, int target) { int left = 0, right = nums.length - 1; // 搜索区间为 [left, right]
while (left <= right) {
int mid = left + (right - left) / 2; if (nums[mid] < target) {
// 搜索区间变为 [mid+1, right]
            left = mid + 1;
        } else if (nums[mid] > target) {
// 搜索区间变为 [left, mid-1]
right = mid - 1;
} else if (nums[mid] == target) {
// 收缩右侧边界
right = mid - 1; }
}
// 检查出界情况
if (left >= nums.length || nums[left] != target) return -1;
    return left;
}

这样就和第一种二分搜索算法统一了,都是两端都闭的「搜索区间」,而且 最后返回的也是 left 变量的值。只要把住二分搜索的逻辑,两种形式大 家看自己喜欢哪种记哪种吧。

Right bound

寻找右侧边界的二分查找 类似寻找左侧边界的算法,这里也会提供两种写法,还是先写常⻅的左闭右 开的写法,只有两处和搜索左侧边界不同,已标注:

int right_bound(int[] nums, int target) { if (nums.length == 0) return -1;
int left = 0, right = nums.length;
    while (left < right) {
        int mid = (left + right) / 2;
        if (nums[mid] == target) {
left = mid + 1; // 注意
} else if (nums[mid] < target) {
            left = mid + 1;
        } else if (nums[mid] > target) {
right = mid; }
}
return left - 1; // 注意 }

为什么这个算法能够找到右侧边界? 答:类似地,关键点还是这里:

if (nums[mid] == target) {
    left = mid + 1;
     nums[mid] == target 不要立即返回而是增大搜索区间的下界 left 使得区间不断向右收缩达到锁定右侧边界的目的

是否也可以把这个算法的「搜索区间」也统一成两端都闭的形式呢?这 样这三个写法就完全统一了,以后就可以闭着眼睛写出来了。 答:当然可以,类似搜索左侧边界的统一写法,其实只要改两个地方就行 了:

int right_bound(int[] nums, int target) { int left = 0, right = nums.length - 1; while (left <= right) {
    int mid = left + (right - left) / 2; if (nums[mid] < target) {
                left = mid + 1;
            } else if (nums[mid] > target) {
    right = mid - 1;
    } else if (nums[mid] == target) {
    // 这里改成收缩左侧边界即可
                left = mid + 1;
            }
    }
    // 这里改为检查 right 越界的情况,⻅下图
    if (right < 0 || nums[right] != target)
    return -1; return right;
    }

Code tips

map.put(key, map.getOrDefault(key, 0) + 1)

Sliding window concepts

滑动窗口算法的思路是这样: 1、我们在字符串 S 中使用双指针中的左右指针技巧,初始化 left = right = 0 ,把索引左闭右开区间 [left, right) 称为一个「窗口」。 2、我们先不断地增加 right 指针扩大窗口 [left, right) ,直到窗口中 的字符串符合要求(包含了 T 中的所有字符)。 3、此时,我们停止增加 right ,转而不断增加 left 指针缩小窗口 [left, right) ,直到窗口中的字符串不再符合要求(不包含 T 中的所有 字符了)。同时,每次增加 left ,我们都要更新一轮结果。 4、重复第 2 和第 3 步,直到 right 到达字符串 S 的尽头。

这个思路其实也不难,第 2 步相当于在寻找一个「可行解」,然后第 3 步在 优化这个「可行解」,最终找到最优解,也就是最短的覆盖子串。左右指针 轮流前进,窗口大小增增减减,窗口不断向右滑动,这就是「滑动窗口」这 个名字的来历。

KPM

Be advised shadow pointer will only work for one case, that is the pattern with repeated char sets same as position zero

// base case,  state will changed "0" -> "1" for given char at 0
  dp[0][pattern.charAt(0)]=1;
  int shaldow=0;
  for (int i = 1; i < n; i++) {
      // to traver each char
      for (int c = 0; c < 256; c++) {
          if(pattern.charAt(i)==c){
              dp[i][c]=i+1;    //[!!!!111] Not = dp[i][c]+1;, should be i+1
          }else{
              dp[i][c]=dp[shaldow][c];
          }
      }
      shaldow=dp[shaldow][pattern.charAt(i)];  // ONLY start from "0" to check with prefix and save time
  }

max profit buy sell stock

V2

Buy and sell stocks without any limitations

public int maxProfit(int[] prices) {
    if(prices==null || prices.length<0)
        return 0;
    int n=prices.length;
    int profitNoholding=0,profitHolding=Integer.MIN_VALUE; // initially no transactions, the holding position will only be invalid
    for(int i=0;i<n;i++){
        int temp = profitNoholding;
        profitNoholding=Math.max(profitNoholding, profitHolding+prices[i]);
        profitHolding = Math.max(profitHolding, temp-prices[i]);
    }

    return profitNoholding;
}


public int maxProfit_best(int[] prices) {
    int total=0;
    // the philosophy is : add to total if current price is greater than previous one
    for(int i=1;i<prices.length;i++){
        if(prices[i]>prices[i-1]){
            total+=prices[i]-prices[i-1];
        }
    }
    return total;
}
}

Next Greater Element

class Solution {
    public int[] nextGreaterElements(int[] nums) {
        if(nums==null) return nums;
        // for next greater , monotonic stack
        Stack<Integer> stack = new Stack();
        int n=nums.length;
        int[] rtn=new int[n];
        // looped, so use "%n" + 2n for loop processing
        for(int i=2*n-1;i>=0;i--){
            // to strip off unqualified element, aka. to keep monotonic stak
            while(!stack.isEmpty() && stack.peek()<=nums[i%n]){ //[!!!!] be careful of bug, it should be "<=", rather than "<", becaust the question is "greater than", and it will add current element to stack, so if "=", it should be populate out as well
                stack.pop();
            }
            rtn[i%n]=stack.isEmpty()?-1:stack.peek();
            stack.push(nums[i%n]);
        }
    return rtn;
    }
}

递归反转整个链表

ListNode reverse(ListNode head) {
if (head.next == null) 
    return head; 
ListNode last = reverse(head.next); 
head.next.next = head;
head.next = null;
return last;
}

对于递归算法,最重要的就是明确递归函数的定义。具体来说,我们的 reverse 函数定义是这样的: 输入一个节点 head ,将「以 head 为起点」的链表反转,并返回反转之 后的头结点。

2022

Linux Tips

Remember, some things have to end for better things to begin.

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2021

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2020

Debug Stuck IntelliJ

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Curl

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2019

Conversations with God

Feelings is the language of the soul. If you want to know what’s true for you about something, look to how your’re feeling about.

Kafka In Spring

Enable Kafka listener annotated endpoints that are created under the covers by a AbstractListenerContainerFactory. To be used on Configuration classes as fol...

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Foreign Exchange

currency pairs Direct ccy: means USD is part of currency pair Cross ccy: means ccy wihtout USD, so except NDF, the deal will be split to legs, both with...

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2018

Guice

A new type of Juice Put simply, Guice alleviates the need for factories and the use of new in your Java code. Think of Guice’s @Inject as the new new. You wi...

YAML

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Cucumber

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JXM

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Strategy-Of-Openshift-Releases

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NodeJs Notes

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CORS :Cross-Origin Resource Sharing

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iOS programming

View A view is also a responder (UIView is a subclass of UIResponder). This means that a view is subject to user interactions, such as taps and swipes. Thus,...

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2017

cloud computering

openshift vs openstack The shoft and direct answer is `OpenShift Origin can run on top of OpenStack. They are complementary projects that work well together....

cloud computering

Concepts Cloud computing is the on-demand demand delivery of compute database storage applications and other IT resources through a cloud services platform v...

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Container

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promise vs observiable

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JDK source

interface RandomAccess Marker interface used by List implementations to indicate that they support fast (generally constant time) random access. The primary ...

SSH SFTP

Secure FTP SFTP over FTP is the equivalant of HTTPS over HTTP, the security version

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After establishing a SSH session, you can install a default web server by executing sudo yum install httpd -y. To start the web server, type sudo service htt...

Oracle

ORA-12899: Value Too Large for Column

Kindle notes

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Java Security Notes

Java Security well-behaved: programs should be prevent from consuming too much system resources

R Language

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Eclipse notes

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Algorithm

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2016

Java GC notes

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Concepts LVS means Linux Virtual Server, which is one Linux built-in component.

Python

(‘—–Unexpected error:’, <type ‘exceptions.TypeError’>) datetime.datetime.now()

Microservices vs. SOA

Microservice Services are organized around capabilities, e.g., user interface front-end, recommendation, logistics, billing, etc. Services are small in ...

Java Class Loader

Codecache The maximum size of the code cache is set via the -XX:ReservedCodeCacheSize=N flag (where N is the default just mentioned for the particular com...

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