feat: add python and java solutions to lcof question

添加《剑指 Offer》题解：面试题20. 表示数值的字符串

Author: yanglbme

lcof/面试题20. 表示数值的字符串/README.md

@@ -8,20 +8,80 @@
 
 ## 解法
 <!-- 这里可写通用的实现逻辑 -->
+遍历字符串：
 
+- 出现 `+`/`-` 时，位置必须是在第 0 位，或者 `e`/`E` 的后面一位
+- 出现 `.` 时，在此之前不能出现 `.` 或者 `e`/`E`
+- 出现 `e`/`E` 时，前面不能出现 `e`/`E`，并且必须出现过数字
 
 ### Python3
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def isNumber(self, s: str) -> bool:
+        if not s or not s.strip():
+            return False
+        s = s.strip()
+        find_num = find_dot = find_e = False
+        for i in range(len(s)):
+            if s[i] == '+' or s[i] == '-':
+                if i != 0 and s[i - 1] != 'e' and s[i - 1] != 'E':
+                    return False
+            elif s[i] >= '0' and s[i] <= '9':
+                find_num = True
+            elif s[i] == '.':
+                if find_dot or find_e:
+                    return False
+                find_dot = True
+            elif s[i] == 'e' or s[i] == 'E':
+                if not find_num or find_e:
+                    return False
+                find_e = True
+                find_num = False
+            else:
+                return False
+        return find_num
 ```
 
 ### Java
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public boolean isNumber(String s) {
+        if (s == null || s.trim().length() == 0) {
+            return false;
+        }
+        char[] chars = s.trim().toCharArray();
+        boolean findNum = false;
+        boolean findE = false;
+        boolean findDot = false;
+        for (int i = 0, n = chars.length; i < n; ++i) {
+            if (chars[i] == '+' || chars[i] == '-') {
+                if (i != 0 && chars[i - 1] != 'e' && chars[i - 1] != 'E') {
+                    return false;
+                }
+            } else if (chars[i] >= '0' && chars[i] <= '9') {
+                findNum = true;
+            } else if (chars[i] == '.') {
+                if (findDot || findE) {
+                    return false;
+                }
+                findDot = true;
+            } else if (chars[i] == 'e' || chars[i] == 'E') {
+                if (findE || !findNum) {
+                    return false;
+                }
+                findE = true;
+                findNum = false; // 确保e之后也出现数
+            } else {
+                return false;
+            }
+        }
+        return findNum;
+    }
+}
 ```
 
 ### ...

lcof/面试题20. 表示数值的字符串/Solution.java

@@ -0,0 +1,34 @@
+class Solution {
+    public boolean isNumber(String s) {
+        if (s == null || s.trim().length() == 0) {
+            return false;
+        }
+        char[] chars = s.trim().toCharArray();
+        boolean findNum = false;
+        boolean findE = false;
+        boolean findDot = false;
+        for (int i = 0, n = chars.length; i < n; ++i) {
+            if (chars[i] == '+' || chars[i] == '-') {
+                if (i != 0 && chars[i - 1] != 'e' && chars[i - 1] != 'E') {
+                    return false;
+                }
+            } else if (chars[i] >= '0' && chars[i] <= '9') {
+                findNum = true;
+            } else if (chars[i] == '.') {
+                if (findDot || findE) {
+                    return false;
+                }
+                findDot = true;
+            } else if (chars[i] == 'e' || chars[i] == 'E') {
+                if (findE || !findNum) {
+                    return false;
+                }
+                findE = true;
+                findNum = false; // 确保e之后也出现数
+            } else {
+                return false;
+            }
+        }
+        return findNum;
+    }
+}

lcof/面试题20. 表示数值的字符串/Solution.py

@@ -0,0 +1,24 @@
+class Solution:
+    def isNumber(self, s: str) -> bool:
+        if not s or not s.strip():
+            return False
+        s = s.strip()
+        find_num = find_dot = find_e = False
+        for i in range(len(s)):
+            if s[i] == '+' or s[i] == '-':
+                if i != 0 and s[i - 1] != 'e' and s[i - 1] != 'E':
+                    return False
+            elif s[i] >= '0' and s[i] <= '9':
+                find_num = True
+            elif s[i] == '.':
+                if find_dot or find_e:
+                    return False
+                find_dot = True
+            elif s[i] == 'e' or s[i] == 'E':
+                if not find_num or find_e:
+                    return False
+                find_e = True
+                find_num = False
+            else:
+                return False
+        return find_num