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feat: add solutions to lc problem: No.0408
No.0408.Valid Word Abbreviation
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‎solution/0200-0299/0277.Find the Celebrity/README.md

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Original file line numberDiff line numberDiff line change
@@ -98,7 +98,7 @@ ans not knows 6
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ans = 6
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```
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$ans$ 认识 $3$，说明 $ans$ 不是名人（$0$ 不是名人），那么名人会是 $1$ 或者 $2$ 吗？不会！因为若 $1$ 或者 $2$ 是名人，那么 $0$ 应该认识 $1$ 或者 $2$ 才对，与前面的例子冲突。因此，我们可以直接将 $ans$ 更新为 $i$。
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这里 $ans$ 认识 $3$，说明 $ans$ 不是名人（$0$ 不是名人），那么名人会是 $1$ 或者 $2$ 吗？不会！因为若 $1$ 或者 $2$ 是名人，那么 $0$ 应该认识 $1$ 或者 $2$ 才对，与前面的例子冲突。因此，我们可以直接将 $ans$ 更新为 $i$。
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我们找出 $ans$ 之后，接下来再遍历一遍，判断 $ans$ 是否满足名人的条件。若不满足，返回 $-1$。
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‎solution/0400-0499/0408.Valid Word Abbreviation/README.md

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@@ -62,22 +62,144 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：模拟**
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模拟字符匹配替换。
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同时遍历 $word$ 和 $abbr$，若 $abbr$ 遇到数字，则 $word$ 跳过对应数字长度的字符数。若数字为空，或者有前导零，则提前返回 false。
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时间复杂度 $O(m+n)$，空间复杂度 $O(1)$。其中 $m$ 是 $word$ 的长度，$n$ 是 $abbr$ 的长度。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def validWordAbbreviation(self, word: str, abbr: str) -> bool:
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i = j = 0
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m, n = len(word), len(abbr)
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while i < m:
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if j >= n:
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return False
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if word[i] == abbr[j]:
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i, j = i + 1, j + 1
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continue
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k = j
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while k < n and abbr[k].isdigit():
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k += 1
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t = abbr[j: k]
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if not t.isdigit() or t[0] == '0' or int(t) == 0:
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return False
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i += int(t)
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j = k
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return i == m and j == n
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public boolean validWordAbbreviation(String word, String abbr) {
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int m = word.length(), n = abbr.length();
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int i = 0, j = 0;
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while (i < m) {
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if (j >= n) {
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return false;
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}
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if (word.charAt(i) == abbr.charAt(j)) {
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++i;
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++j;
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continue;
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}
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int k = j;
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while (k < n && Character.isDigit(abbr.charAt(k))) {
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++k;
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}
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String t = abbr.substring(j, k);
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if (j == k || t.charAt(0) == '0' || Integer.parseInt(t) == 0) {
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return false;
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}
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i += Integer.parseInt(t);
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j = k;
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}
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return i == m && j == n;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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bool validWordAbbreviation(string word, string abbr) {
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int i = 0, j = 0;
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int m = word.size(), n = abbr.size();
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while (i < m) {
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if (j >= n) {
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return false;
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}
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if (word[i] == abbr[j]) {
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++i;
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++j;
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continue;
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}
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int k = j;
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while (k < n && isdigit(abbr[k])) {
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++k;
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}
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string t = abbr.substr(j, k - j);
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if (k == j || t[0] == '0') {
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return false;
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}
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int x = stoi(t);
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if (x == 0) {
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return false;
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}
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i += x;
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j = k;
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}
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return i == m && j == n;
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}
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};
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```
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### **Go**
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```go
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func validWordAbbreviation(word string, abbr string) bool {
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i, j := 0, 0
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m, n := len(word), len(abbr)
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for i < m {
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if j >= n {
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return false
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}
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if word[i] == abbr[j] {
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i++
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j++
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continue
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}
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k := j
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for k < n && abbr[k] >= '0' && abbr[k] <= '9' {
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k++
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}
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if k == j || abbr[j] == '0' {
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return false
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}
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x, _ := strconv.Atoi(abbr[j:k])
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if x == 0 {
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return false
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}
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i += x
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j = k
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}
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return i == m && j == n
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}
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```
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### **...**

‎solution/0400-0499/0408.Valid Word Abbreviation/README_EN.md

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@@ -63,13 +63,127 @@
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### **Python3**
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```python
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class Solution:
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def validWordAbbreviation(self, word: str, abbr: str) -> bool:
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i = j = 0
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m, n = len(word), len(abbr)
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while i < m:
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if j >= n:
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return False
73+
if word[i] == abbr[j]:
74+
i, j = i + 1, j + 1
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continue
76+
k = j
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while k < n and abbr[k].isdigit():
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k += 1
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t = abbr[j: k]
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if not t.isdigit() or t[0] == '0' or int(t) == 0:
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return False
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i += int(t)
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j = k
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return i == m and j == n
6785
```
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### **Java**
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```java
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class Solution {
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public boolean validWordAbbreviation(String word, String abbr) {
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int m = word.length(), n = abbr.length();
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int i = 0, j = 0;
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while (i < m) {
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if (j >= n) {
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return false;
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}
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if (word.charAt(i) == abbr.charAt(j)) {
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++i;
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++j;
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continue;
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}
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int k = j;
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while (k < n && Character.isDigit(abbr.charAt(k))) {
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++k;
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}
107+
String t = abbr.substring(j, k);
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if (j == k || t.charAt(0) == '0' || Integer.parseInt(t) == 0) {
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return false;
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}
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i += Integer.parseInt(t);
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j = k;
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}
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return i == m && j == n;
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}
116+
}
117+
```
118+
119+
### **C++**
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```cpp
122+
class Solution {
123+
public:
124+
bool validWordAbbreviation(string word, string abbr) {
125+
int i = 0, j = 0;
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int m = word.size(), n = abbr.size();
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while (i < m) {
128+
if (j >= n) {
129+
return false;
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}
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if (word[i] == abbr[j]) {
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++i;
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++j;
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continue;
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}
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int k = j;
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while (k < n && isdigit(abbr[k])) {
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++k;
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}
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string t = abbr.substr(j, k - j);
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if (k == j || t[0] == '0') {
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return false;
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}
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int x = stoi(t);
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if (x == 0) {
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return false;
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}
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i += x;
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j = k;
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}
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return i == m && j == n;
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}
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};
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```
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### **Go**
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```go
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func validWordAbbreviation(word string, abbr string) bool {
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i, j := 0, 0
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m, n := len(word), len(abbr)
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for i < m {
163+
if j >= n {
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return false
165+
}
166+
if word[i] == abbr[j] {
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i++
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j++
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continue
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}
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k := j
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for k < n && abbr[k] >= '0' && abbr[k] <= '9' {
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k++
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}
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if k == j || abbr[j] == '0' {
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return false
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}
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x, _ := strconv.Atoi(abbr[j:k])
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if x == 0 {
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return false
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}
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i += x
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j = k
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}
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return i == m && j == n
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}
73187
```
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### **...**

‎solution/0400-0499/0408.Valid Word Abbreviation/Solution.cpp

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,32 @@
1+
class Solution {
2+
public:
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bool validWordAbbreviation(string word, string abbr) {
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int i = 0, j = 0;
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int m = word.size(), n = abbr.size();
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while (i < m) {
7+
if (j >= n) {
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return false;
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}
10+
if (word[i] == abbr[j]) {
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++i;
12+
++j;
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continue;
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}
15+
int k = j;
16+
while (k < n && isdigit(abbr[k])) {
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++k;
18+
}
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string t = abbr.substr(j, k - j);
20+
if (k == j || t[0] == '0') {
21+
return false;
22+
}
23+
int x = stoi(t);
24+
if (x == 0) {
25+
return false;
26+
}
27+
i += x;
28+
j = k;
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}
30+
return i == m && j == n;
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}
32+
};

‎solution/0400-0499/0408.Valid Word Abbreviation/Solution.go

+28
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,28 @@
1+
func validWordAbbreviation(word string, abbr string) bool {
2+
i, j := 0, 0
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m, n := len(word), len(abbr)
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for i < m {
5+
if j >= n {
6+
return false
7+
}
8+
if word[i] == abbr[j] {
9+
i++
10+
j++
11+
continue
12+
}
13+
k := j
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for k < n && abbr[k] >= '0' && abbr[k] <= '9' {
15+
k++
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}
17+
if k == j || abbr[j] == '0' {
18+
return false
19+
}
20+
x, _ := strconv.Atoi(abbr[j:k])
21+
if x == 0 {
22+
return false
23+
}
24+
i += x
25+
j = k
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}
27+
return i == m && j == n
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}

‎solution/0400-0499/0408.Valid Word Abbreviation/Solution.java

+27
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,27 @@
1+
class Solution {
2+
public boolean validWordAbbreviation(String word, String abbr) {
3+
int m = word.length(), n = abbr.length();
4+
int i = 0, j = 0;
5+
while (i < m) {
6+
if (j >= n) {
7+
return false;
8+
}
9+
if (word.charAt(i) == abbr.charAt(j)) {
10+
++i;
11+
++j;
12+
continue;
13+
}
14+
int k = j;
15+
while (k < n && Character.isDigit(abbr.charAt(k))) {
16+
++k;
17+
}
18+
String t = abbr.substring(j, k);
19+
if (j == k || t.charAt(0) == '0' || Integer.parseInt(t) == 0) {
20+
return false;
21+
}
22+
i += Integer.parseInt(t);
23+
j = k;
24+
}
25+
return i == m && j == n;
26+
}
27+
}

‎solution/0400-0499/0408.Valid Word Abbreviation/Solution.py

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,19 @@
1+
class Solution:
2+
def validWordAbbreviation(self, word: str, abbr: str) -> bool:
3+
i = j = 0
4+
m, n = len(word), len(abbr)
5+
while i < m:
6+
if j >= n:
7+
return False
8+
if word[i] == abbr[j]:
9+
i, j = i + 1, j + 1
10+
continue
11+
k = j
12+
while k < n and abbr[k].isdigit():
13+
k += 1
14+
t = abbr[j: k]
15+
if not t.isdigit() or t[0] == '0' or int(t) == 0:
16+
return False
17+
i += int(t)
18+
j = k
19+
return i == m and j == n

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