Algorithm Notes
  • Introduction
  • Search & Backtracking 搜索与回溯
    • Tree 与 BackTracking 的比较
    • Subsets, Combination 与 Permutation
    • Subsets & Combinations & Combination Sum
    • 枚举法
    • N 皇后 + 矩阵 Index Trick
    • Sudoku 数独 + 矩阵 Index Trick
    • Word Ladder I & II
    • Number of ways 类
    • DFS flood filling
    • Strobogrammatic 数生成
    • String 构造式 DFS + Backtracking
    • Word Pattern I & II
    • (G) Binary Watch
    • (FB) Phone Letter Combination
    • 常见搜索问题的迭代解法
  • String,字符串类
    • 多步翻转法
    • Substring 结构和遍历
    • Palindrome 问题
    • Palindrome Continued
    • String / LinkedList 大数运算
    • 序列化与压缩
    • 5/24 String 杂题
    • Knuth–Morris–Pratt 字符串匹配
    • Lempel–Ziv–Welch 字符串压缩算法
    • (G) Decode String
    • (G) UTF-8 Validation
  • Binary Tree,二叉树
    • 各种 Binary Tree 定义
    • LCA 类问题
    • 三序遍历,vertical order
    • Post order traversal 的应用
    • Min/Max/Balanced Depth
    • BST
    • 子树结构
    • Level Order traversal
    • Morris 遍历
    • 修改结构
    • 创建 / 序列化
    • 子树组合,BST query
    • 路径与路径和
    • NestedInteger 类
    • (FB) 从 Binary Tree Path 看如何递归转迭代
    • (FB) Binary Tree Path 比较路径大小
    • 比较好玩的 Binary Tree 概率题
  • Segment & Fenwick Tree,区间树
    • Segment Tree 基础操作
    • Segment Tree 的应用
    • Fenwick Tree (Binary Indexed Tree)
    • Range Sum Query 2D - Immutable
  • Union-Find,并查集
    • Union-Find,并查集基础
    • Union-Find, 并查集应用
  • Dynamic Programming, 动态规划
    • 6/20, 入门 House Robber
    • 7/12, Paint Fence / House
    • 6/24, 滚动数组
    • 6/24, 记忆化搜索
    • 6/24, 博弈类 DP
    • 博弈类DP, Flip Game
    • 6/25, 区间类DP
    • 6/27, subarray 划分类,股票
    • 7/2, 字符串类
    • Bomb Enemies
    • 8/2,背包问题
    • (G) Max Vacation
    • (11/4新增) AST 子树结构 DP
  • LinkedList,链表
    • 6/9, LinkedList,反转与删除
    • 6/11, LinkedList 杂题
    • (FB) 链表的递归与倒序打印
  • LinkedIn 面经,算法题
    • 6/17, LinkedIn 面经题
    • 6/28, LinkedIn 面经题
    • 7/6, LinkedIn 面经
    • Shortest Word Distance 类
    • DFA Parse Integer
  • Two Pointers,双指针
    • 3 Sum, 3 Sum Closest / Smaller, 4 Sum
    • 对撞型,灌水类
    • 对撞型,partition类
    • Wiggle Sort I & II
    • 双指针,窗口类
    • 双指针,窗口类
    • Heap,排序 matrix 中的 two pointers
  • Bit & Math,位运算与数学
    • Bit Manipulation,对于 '1' 位的操作
    • Math & Bit Manipulation, Power of X
    • 坐标系 & 数值计算类
    • Add Digits
    • 用 int 做字符串 signature
  • Interval 与 扫描线
    • Range Addition & LCS
    • 7/5, Interval 类,扫描线
  • Trie,字典树
    • 6/9, Trie, 字典树
  • 单调栈,LIS
    • 4/13 LIS
    • 栈, 单调栈
    • Largest Divisible Subset
  • Binary Search 类
    • Matrix Binary Search
    • Array Binary Search
    • Find Peak Element I & II
    • **Median of Two Sorted Arrays
  • Graph & Topological Sort,图 & 拓扑排序
    • 有向 / 无向 图的基本性质和操作
    • 拓扑排序, DFS 做法
    • 拓扑排序, BFS 做法
    • Course Schedule I & II
    • Alien Dictionary
    • Undirected Graph, BFS
    • Undirected Graph, DFS
    • 矩阵,BFS 最短距离探索
    • 欧拉回路,Hierholzer算法
    • AI, 迷宫生成
    • AI, 迷宫寻路算法
    • (G) Deep Copy 无向图成有向图
  • 括号与数学表达式的计算
  • Iterator 类
  • Majority Element,Moore's Voting
  • Matrix Inplace Operations
  • 常见数据结构设计
  • (G) Design / OOD 类算法题
  • 随机算法 & 数据结构
  • (FB) I/O Buffer
  • (FB) Simplify Path, H-Index I & II
  • (FB) Excel Sheet, Remove Duplicates
  • Integer 的构造,操作,序列化
  • Frequency 类问题
  • Missing Number 类,元素交换,数组环形跳转
  • 8/10, Google Tag
  • (FB) Rearrange String k Distance Apart
  • Abstract Algebra
    • Chap1 -- Why Abstract Algebra ?
    • Chap2 -- Operations
    • Chap3 -- The Definition of Groups
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  • Strobogrammatic Number II
  • 注意:index == 0 并且 i == 0 的时候要跳过,免得在起始位置填上 0 .
  • Strobogrammatic Number III
  • 后来发现有点多余,可以直接用内置的 str1.compareTo(str2).

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  1. Search & Backtracking 搜索与回溯

Strobogrammatic 数生成

PreviousDFS flood fillingNextString 构造式 DFS + Backtracking

Last updated 4 years ago

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为什么一个这么简单的 DFS 能超过 89% ..

注意:index == 0 并且 i == 0 的时候要跳过,免得在起始位置填上 0 .

public class Solution {
    public List<String> findStrobogrammatic(int n) {
        List<String> list = new ArrayList<>();
        char[] num1 = {'0','1','8','6','9'};
        char[] num2 = {'0','1','8','9','6'};
        char[] number = new char[n];

        dfs(list, number, num1, num2, 0);

        return list;
    }

    private void dfs(List<String> list, char[] number, char[] num1, char[] num2, int index){
        int left = index;
        int right = number.length - index - 1;

        if(left > right){
            list.add(new String(number));
            return;
        }
        // We can fill in 0,1,8 only
        if(left == right){
            for(int i = 0; i < 3; i++){
                number[left] = num1[i];
                dfs(list, number, num1, num2, index + 1);
            }
        } else {
            for(int i = 0; i < num1.length; i++){
                if(index == 0 && i == 0) continue;
                number[left] = num1[i];
                number[right] = num2[i];
                dfs(list, number, num1, num2, index + 1);
            }
        }
    }
}

Google 面经里的 follow-up 是,给定一个上限 n ,输出所有上限范围内的数。

办法土了点,遍历所有 lowLen ~ highLen 区间的长度,生成所有可能的结果,考虑到区间可能是大数,我们就改一下,自己写一个 String compare 函数好了。

后来发现有点多余,可以直接用内置的 str1.compareTo(str2).

超过 81.92% ~

public class Solution {
    int count = 0;
    public int strobogrammaticInRange(String low, String high) {
        int lowLen = low.length();
        int highLen = high.length();

        char[] num1 = {'0','1','8','6','9'};
        char[] num2 = {'0','1','8','9','6'};

        for(int i = lowLen; i <= highLen; i++){
            char[] number = new char[i];
            dfs(number, num1, num2, 0, low, high);
        }

        return count;
    }

    private void dfs(char[] number, char[] num1, char[] num2, int index, String low, String high){
        int left = index;
        int right = number.length - index - 1;

        if(left > right){
            String num = new String(number);
            if(compare(low, num) <= 0 && compare(num, high) <= 0) count++;
            return;
        } else if(left == right){
            for(int i = 0; i < 3; i++){
                number[left] = num1[i];
                dfs(number, num1, num2, index + 1, low, high);
            }
        } else {
            for(int i = 0; i < 5; i++){
                if(index == 0 && i == 0) continue;
                number[left] = num1[i];
                number[right] = num2[i];
                dfs(number, num1, num2, index + 1, low, high);
            }
        }
    }

    // -1 : str1 is bigger
    // 1 : str 2 is bigger
    // 0 : equal
    private int compare(String str1, String str2){
        if(str1.length() > str2.length()) return 1;
        else if(str1.length() < str2.length()) return -1;
        else {
            for(int i = 0; i < str1.length(); i++){
                int digit1 = str1.charAt(i) - '0';
                int digit2 = str2.charAt(i) - '0';

                if(digit1 != digit2) return (digit1 > digit2) ? 1: -1;
            }
        }
        // Equal
        return 0;
    }

}

Strobogrammatic Number II
Strobogrammatic Number III