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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  • Integer to English Words
  • Roman to Integer
  • Count and Say
  • Integer to Roman
  • Integer to English Words

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Integer 的构造,操作,序列化

Previous(FB) Excel Sheet, Remove DuplicatesNextFrequency 类问题

Last updated 4 years ago

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  • 小于 20 的数要字典;

  • 十几 Tens 的,需要字典;

  • 多少个 thousands 的,需要字典,从右往左 index 递增;

  • 以三位数为单位处理,任何三位数都可以用 helper function + 字典解决,自带 hundred 单位。

  • 0 在所有情况都代表空字符,除了 num 一开始就等于 0 的情况要返回 "Zero".

自己第一遍 AC 的版本太粗糙,就不放了。

这个版本就简洁了很多,从右向左,递归调用处理三位数的情况;

public class Solution {
    private final String[] LESS_THAN_20 = {"", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"};
    private final String[] TENS = {"", "Ten", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"};
    private final String[] THOUSANDS = {"", "Thousand", "Million", "Billion"};

    public String numberToWords(int num) {
        if(num == 0) return "Zero";
        String rst = "";
        int highPtr = 0;
        while(num != 0){
            if(num % 1000 != 0){
                rst = helper(num % 1000) + THOUSANDS[highPtr] + " " + rst;
            }
            num /= 1000;
            highPtr ++;
        }
        return rst.trim();
    }

    private String helper(int num){
        if(num == 0)
            return "";
        else if(num < 20)
            return LESS_THAN_20[num] + " ";
        else if(num < 100)
            return TENS[num / 10] + " " + helper(num % 10);
        else 
            return LESS_THAN_20[num / 100] + ' ' + "Hundred" + ' ' + helper(num % 100);
    }
}

Trivial problem. 没啥好讲的。。

public class Solution {
    public int romanToInt(String s) {
        if(s.length() == 0) return 0;

        int num = getNum(s.charAt(0));
        int prev = num;
        for(int i = 1; i < s.length(); i++){
            int cur = getNum(s.charAt(i));
            if(cur > prev){
                num -= 2 * prev;
            }
            prev = cur;
            num += cur;
        }

        return num;
    }

    private int getNum(char chr){
        switch(chr){
            case 'I':
                return 1;
            case 'V':
                return 5;
            case 'X':
                return 10;
            case 'L':
                return 50;
            case 'C':
                return 100;
            case 'D':
                return 500;
            case 'M':
                return 1000;                
            default:
                return 0;
        }
    }
}

这题怪怪的,我也不知道到底想考啥。。难道是迭代和递归么。。。。

public class Solution {
    public String countAndSay(int n) {
        if(n == 1) return "1";
        StringBuilder sb = new StringBuilder();
        sb.append("1");

        for(int i = 1; i < n; i++){
            String str = sb.toString();
            sb.setLength(0);
            int index = 0;
            while(index < str.length()){
                int num = str.charAt(index) - '0';
                int count = 1;
                while(index < str.length() - 1 && str.charAt(index) == str.charAt(index + 1)){
                    count ++;
                    index ++;
                }
                index ++;
                sb.append(count);
                sb.append(num);
            }
        }

        return sb.toString();
    }
}

之前那道 Roman to Integer 就弄了个 switch case ,这次可能性稍微多了点,直接开两个 1-1 onto mapping 当表查好了。

  • 当可能的情况“有限”并“可数”的时候,可以自己用 array 去建 1-1 mapping 便于查询。

public class Solution {
    public String intToRoman(int num) {
        int[] nums =      {1000,  900,  500,  400,  100,   90,  50,   40,   10,    9,   5,    4,   1};
        String[] romans = { "M", "CM",  "D", "CD",  "C", "XC", "L", "XL",  "X", "IX", "V", "IV", "I"};

        StringBuilder sb = new StringBuilder();
        while(num > 0){
            for(int i = 0; i < nums.length; i++){
                if(num >= nums[i]){
                    num -= nums[i];
                    sb.append(romans[i]);
                    break;
                }
            }
        }

        return sb.toString();
    }
}
  • 小于 20 的数要字典;

  • 十几 Tens 的,需要字典;

  • 多少个 thousands 的,需要字典,从右往左 index 递增;

  • 以三位数为单位处理,任何三位数都可以用 helper function + 字典解决,自带 hundred 单位。

  • 0 在所有情况都代表空字符,除了 num 一开始就等于 0 的情况要返回 "Zero".

自己第一遍 AC 的版本太粗糙,就不放了。

这个版本就简洁了很多,从右向左,递归调用处理三位数的情况;

public class Solution {
    private final String[] LESS_THAN_20 = {"", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"};
    private final String[] TENS = {"", "Ten", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"};
    private final String[] THOUSANDS = {"", "Thousand", "Million", "Billion"};

    public String numberToWords(int num) {
        if(num == 0) return "Zero";
        String rst = "";
        int highPtr = 0;
        while(num != 0){
            if(num % 1000 != 0){
                rst = helper(num % 1000) + THOUSANDS[highPtr] + " " + rst;
            }
            num /= 1000;
            highPtr ++;
        }
        return rst.trim();
    }

    private String helper(int num){
        if(num == 0)
            return "";
        else if(num < 20)
            return LESS_THAN_20[num] + " ";
        else if(num < 100)
            return TENS[num / 10] + " " + helper(num % 10);
        else 
            return LESS_THAN_20[num / 100] + ' ' + "Hundred" + ' ' + helper(num % 100);
    }
}

Integer to English Words
Roman to Integer
Count and Say
Integer to Roman
Integer to English Words