[LeetCode] 1805. Number of Different Integers in a String

You are given a string word that consists of digits and lowercase English letters.

You will replace every non-digit character with a space. For example, "a123bc34d8ef34" will become " 123  34 8  34". Notice that you are left with some integers that are separated by at least one space: "123", "34", "8", and "34".

Return the number of different integers after performing the replacement operations on word.

Two integers are considered different if their decimal representations without any leading zeros are different.

Example 1:

Input: word = "a123bc34d8ef34"
Output: 3
Explanation: The three different integers are "123", "34", and "8". Notice that "34" is only counted once.

Example 2:

Input: word = "leet1234code234"
Output: 2

Example 3:

Input: word = "a1b01c001"
Output: 1
Explanation: The three integers "1", "01", and "001" all represent the same integer because
the leading zeros are ignored when comparing their decimal values.

Constraints:

• 1 <= word.length <= 1000
• word consists of digits and lowercase English letters.

字符串中不同整数的数目。

给你一个字符串 word ，该字符串由数字和小写英文字母组成。

请你用空格替换每个不是数字的字符。例如，"a123bc34d8ef34" 将会变成 " 123  34 8  34" 。注意，剩下的这些整数为（相邻彼此至少有一个空格隔开）："123"、"34"、"8" 和 "34" 。

返回对 word 完成替换后形成的 不同 整数的数目。

只有当两个整数的 不含前导零 的十进制表示不同， 才认为这两个整数也不同。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/number-of-different-integers-in-a-string
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路是双指针。这道题是简单题，但是有一些 corner case 需要处理。题意不难理解，就是把 input 字符串里的数字都挑出来，然后返回 unique 的数字的个数。由于需要返回的只是数字的个数，所以我们不需要把字符串转换成数字，这样也就没有整型溢出的问题了。corner case 是比如例子三，对于1, 01, 001，我们都视为 1。

时间O(n)

空间O(n)

Java实现

 1 class Solution {
 2     public int numDifferentIntegers(String word) {
 3         // corner case
 4         if (word == null || word.length() == 0) {
 5             return 0;
 6         }
 7         
 8         // normal case
 9         HashSet<String> set = new HashSet<>();
10         int i = 0;
11         while (i < word.length()) {
12             if (Character.isDigit(word.charAt(i))) {
13                 int j = i;
14                 StringBuilder sb = new StringBuilder();
15                 while (i < word.length() && Character.isDigit(word.charAt(i))) {
16                     sb.append(word.charAt(i));
17                     i++;
18                 }
19                 while (j < i && word.charAt(j) == '0') {
20                     j++;
21                 }
22                 set.add(word.substring(j, i));
23             } else {
24                 i++;
25             }
26         }
27         return set.size();
28     }
29 }

LeetCode 题目总结