README.md

423. 从英文中重建数字

English Version

题目描述

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：

输入：s = "owoztneoer"
输出："012"

示例 2：

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

解法

统计 ["e","g","f","i","h","o","n","s","r","u","t","w","v","x","z"] 每个字母在哪些数字出现过。

字母	数字
e	0 1 3 5 7 8 9
g	8
f	4 5
i	5 6 8 9
h	3 8
o	0 1 2 4
n	1 7 9
s	6 7
r	0 3 4
u	4
t	2 3 8
w	2
v	5 7
x	6
z	0

由于部分字母只在某个数字出现过，比如字母 z 只在 0 出现过，因此我们统计英文中 z 的数量，就可以推断数字 0 的个数，依次类推。

Python3

class Solution:
    def originalDigits(self, s: str) -> str:
        counter = Counter(s)
        cnt = [0] * 10

        cnt[0] = counter['z']
        cnt[2] = counter['w']
        cnt[4] = counter['u']
        cnt[6] = counter['x']
        cnt[8] = counter['g']

        cnt[3] = counter['h'] - cnt[8]
        cnt[5] = counter['f'] - cnt[4]
        cnt[7] = counter['s'] - cnt[6]

        cnt[1] = counter['o'] - cnt[0] - cnt[2] - cnt[4]
        cnt[9] = counter['i'] - cnt[5] - cnt[6] - cnt[8]

        return ''.join(cnt[i] * str(i) for i in range(10))

Java

class Solution {
    public String originalDigits(String s) {
        int[] counter = new int[26];
        for (char c : s.toCharArray()) {
            ++counter[c - 'a'];
        }
        int[] cnt = new int[10];
        cnt[0] = counter['z' - 'a'];
        cnt[2] = counter['w' - 'a'];
        cnt[4] = counter['u' - 'a'];
        cnt[6] = counter['x' - 'a'];
        cnt[8] = counter['g' - 'a'];

        cnt[3] = counter['h' - 'a'] - cnt[8];
        cnt[5] = counter['f' - 'a'] - cnt[4];
        cnt[7] = counter['s' - 'a'] - cnt[6];

        cnt[1] = counter['o' - 'a'] - cnt[0] - cnt[2] - cnt[4];
        cnt[9] = counter['i' - 'a'] - cnt[5] - cnt[6] - cnt[8];

        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < 10; ++i) {
            for (int j = 0; j < cnt[i]; ++j) {
                sb.append(i);
            }
        }
        return sb.toString();
    }
}

C++

class Solution {
public:
    string originalDigits(string s) {
        vector<int> counter(26);
        for (char c : s) ++counter[c - 'a'];
        vector<int> cnt(10);
        cnt[0] = counter['z' - 'a'];
        cnt[2] = counter['w' - 'a'];
        cnt[4] = counter['u' - 'a'];
        cnt[6] = counter['x' - 'a'];
        cnt[8] = counter['g' - 'a'];

        cnt[3] = counter['h' - 'a'] - cnt[8];
        cnt[5] = counter['f' - 'a'] - cnt[4];
        cnt[7] = counter['s' - 'a'] - cnt[6];

        cnt[1] = counter['o' - 'a'] - cnt[0] - cnt[2] - cnt[4];
        cnt[9] = counter['i' - 'a'] - cnt[5] - cnt[6] - cnt[8];

        string ans;
        for (int i = 0; i < 10; ++i)
            for (int j = 0; j < cnt[i]; ++j)
                ans += char(i + '0');
        return ans;
    }
};

Go

func originalDigits(s string) string {
	counter := make([]int, 26)
	for _, c := range s {
		counter[c-'a']++
	}
	cnt := make([]int, 10)
	cnt[0] = counter['z'-'a']
	cnt[2] = counter['w'-'a']
	cnt[4] = counter['u'-'a']
	cnt[6] = counter['x'-'a']
	cnt[8] = counter['g'-'a']

	cnt[3] = counter['h'-'a'] - cnt[8]
	cnt[5] = counter['f'-'a'] - cnt[4]
	cnt[7] = counter['s'-'a'] - cnt[6]

	cnt[1] = counter['o'-'a'] - cnt[0] - cnt[2] - cnt[4]
	cnt[9] = counter['i'-'a'] - cnt[5] - cnt[6] - cnt[8]

	ans := []byte{}
	for i, c := range cnt {
		ans = append(ans, bytes.Repeat([]byte{byte('0' + i)}, c)...)
	}
	return string(ans)
}

...