LeetCode423-从英文中重建数字

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题目

给你一个字符串 s ，其中包含字母顺序打乱的用英文单词表示的若干数字（0-9）。按 升序 返回原始的数字。

示例 1：

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输入：s = "owoztneoer"
输出："012"

示例 2：

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输入：s = "fviefuro"
输出："45"

提示：

  • 1 <= s.length <= 105
  • s[i] 为 [“e”,”g”,”f”,”i”,”h”,”o”,”n”,”s”,”r”,”u”,”t”,”w”,”v”,”x”,”z”] 这些字符之一
  • s 保证是一个符合题目要求的字符串

解答

看完题目后，大概猜出可能需要每个数字的特定字母做标识，所以干脆列出0-9数字的英文单词观察

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public static final String ZERO = "zero";//Z
public static final String ONE = "one";//Z,W,U->O
public static final String TWO = "two";//W
public static final String THREE = "three";//Z,U->R
public static final String FOUR = "four";//U
public static final String FIVE = "five";//U->F
public static final String SIX = "six";//X
public static final String SEVEN = "seven";//X->S
public static final String EIGHT = "eight";//G
public static final String NINE = "nine";//G,X,U->F->I

如上，观察后得出，可以分三轮，计算出每个数字的出现次数

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//第一轮，选出zero,two,four,six,eight
//第二轮，选出one,three,five,seven
//第三轮，选出nine

得出方法后，代码就很容易写了

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import java.util.HashMap;
import java.util.Map;
import java.util.Optional;

class Solution {
    public String originalDigits(String s) {
        //首先把s做分解，用Map<Character, Integer>做容器
        Map<Character, Integer> wordCount = new HashMap<>();
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if( wordCount.containsKey(c) ){
                wordCount.put(c, wordCount.get(c)+1);
            } else {
                wordCount.put(c, 1);
            }
        }

        int[] numCount = new int[10];
        //第一轮
        numCount[0] = Optional.ofNullable(wordCount.get('z')).orElse(0);
        numCount[2] = Optional.ofNullable(wordCount.get('w')).orElse(0);
        numCount[4] = Optional.ofNullable(wordCount.get('u')).orElse(0);
        numCount[6] = Optional.ofNullable(wordCount.get('x')).orElse(0);
        numCount[8] = Optional.ofNullable(wordCount.get('g')).orElse(0);
        //第二轮
        numCount[1] = Optional.ofNullable(wordCount.get('o')).orElse(0)-numCount[0]-numCount[2]-numCount[4];
        numCount[3] = Optional.ofNullable(wordCount.get('r')).orElse(0)-numCount[0]-numCount[4];
        numCount[5] = Optional.ofNullable(wordCount.get('f')).orElse(0)-numCount[4];
        numCount[7] = Optional.ofNullable(wordCount.get('s')).orElse(0)-numCount[6];
        //第三轮
        numCount[9] = Optional.ofNullable(wordCount.get('i')).orElse(0)-numCount[5]-numCount[6]-numCount[8];

        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < numCount.length; i++) {
            for (int j = 0; j < numCount[i]; j++) {
                sb.append(i);
            }
        }
        return sb.toString();
    }
}

http://blog-image-creasylai.oss-cn-shenzhen.aliyuncs.com/blog.images/img/2021/1124/article2/01.png

题目转载自：https://leetcode-cn.com/problems/reconstruct-original-digits-from-english

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