style: code optimization

leetcode-docs/easy/13-roman-to-int.mdx

@@ -115,7 +115,6 @@ pub fn roman_to_int(s: String) -> i32 {
         })
         .0
 }
-
 ```
 
 </TabItem>

leetcode-docs/easy/14-longest-common-prefix.mdx

@@ -25,8 +25,7 @@ sidebar_label: 14. 最长公共前缀
 
 ## 题解
 
-先假设第一个单词为 `prefix`, 如果 `strs` 里所有单词的前缀都是 `prefix`, 那就返回之; 
-否则将 `prefix` 的最后一位删掉继续循环, 直到找出前缀.
+先假设第一个单词为 `prefix`, 如果 `strs` 里所有单词的前缀都是 `prefix`, 那就返回之; 否则将 `prefix` 的最后一位删掉继续循环, 直到找出前缀.
 
 import Tabs from '@theme/Tabs'
 import TabItem from '@theme/TabItem'
@@ -40,13 +39,14 @@ import TabItem from '@theme/TabItem'
  * @return {string}
  */
 var longestCommonPrefix = function (strs) {
+  strs.sort((a, b) => a.length - b.length)
   let prefix = strs[0]
-  let hasFound = false
+  let flag = false
 
-  while (!hasFound) {
-    hasFound = strs.every((str) => str.startsWith(prefix))
+  while (!flag) {
+    flag = strs.every((str) => str.startsWith(prefix))
 
-    if (!hasFound) {
+    if (!flag) {
       prefix = prefix.slice(0, -1)
     }
   }
@@ -60,6 +60,8 @@ var longestCommonPrefix = function (strs) {
 
 ```rust
 pub fn longest_common_prefix(strs: Vec<String>) -> String {
+    let mut strs = strs;
+    strs.sort();
     let mut prefix: &str = &strs[0];
     let mut has_found = false;
 

leetcode-docs/easy/20-is-valid.mdx

@@ -57,7 +57,7 @@ var isValid = function (s) {
   const n = s.length
 
   // 如果是奇数, 一定不是合法的括号
-  if (n % 2 == 1) return false
+  if (n % 2 === 1) return false
 
   const stack = []
 

leetcode-docs/hard/23-merge-k-lists.mdx

@@ -125,6 +125,7 @@ var mergeKLists = function (lists) {
  * @param {ListNode} l2
  * @return {ListNode}
  */
+// 这个就是 21 题原题
 var mergeTwoLists = function (l1, l2) {
   const dummy = new ListNode(-1)
   let curr = dummy

leetcode-docs/hard/76-min-window.mdx

@@ -65,59 +65,42 @@ import TabItem from '@theme/TabItem'
  * @return {string}
  */
 var minWindow = function (s, t) {
-  let res = ''
+  let minStr = ''
 
   const needMap = new Map()
-  for (const letter of t) {
-    if (needMap.has(letter)) {
-      needMap.set(letter, needMap.get(letter) + 1)
-    } else {
-      needMap.set(letter, 1)
-    }
+  for (const ch of t) {
+    needMap.set(ch, needMap.has(ch) ? needMap.get(ch) + 1 : 1)
   }
+
   const windowMap = new Map()
   let start = 0
   let end = 0
   let count = 0
 
-  while (end < sLen) {
-    const endLetter = s[end]
-    end++
-
-    if (needMap.has(endLetter)) {
-      if (windowMap.has(endLetter)) {
-        windowMap.set(endLetter, windowMap.get(endLetter) + 1)
-      } else {
-        windowMap.set(endLetter, 1)
-      }
+  while (end < s.length) {
+    const endCh = s[end++]
 
-      if (needMap.get(endLetter) === windowMap.get(endLetter)) {
-        count++
-      }
+    if (needMap.has(endCh)) {
+      windowMap.set(endCh, windowMap.has(endCh) ? windowMap.get(endCh) + 1 : 1)
+      if (needMap.get(endCh) === windowMap.get(endCh)) count++
     }
 
     // 当 count ===  needMap.size, 说明找到了一个覆盖子串
     // 这个时候就可以收缩窗口了
     while (count === needMap.size) {
-      const window = s.slice(start, end)
-      if (res.length === 0 || window.length < res.length) {
-        res = window
-      }
-
-      const startLetter = s[start]
-      start++
+      const subStr = s.slice(start, end)
+      if (minStr.length === 0 || subStr.length < minStr.length) minStr = subStr
 
-      if (needMap.has(startLetter)) {
-        if (needMap.get(startLetter) === windowMap.get(startLetter)) {
-          count--
-        }
+      const startCh = s[start++]
 
-        windowMap.set(startLetter, windowMap.get(startLetter) - 1)
+      if (needMap.has(startCh)) {
+        if (needMap.get(startCh) === windowMap.get(startCh)) count--
+        windowMap.set(startCh, windowMap.get(startCh) - 1)
       }
     }
   }
 
-  return res
+  return minStr
 }
 ```
 
@@ -128,7 +111,7 @@ var minWindow = function (s, t) {
 use std::collections::HashMap;
 
 pub fn min_window(s: String, t: String) -> String {
-    let mut res = "";
+    let mut min_str = "";
 
     let mut need_map = HashMap::new();
     for letter in t.as_bytes() {
@@ -152,9 +135,9 @@ pub fn min_window(s: String, t: String) -> String {
         }
 
         while count == need_map.len() {
-            let window = &s[start..end];
-            if res.len() == 0 || window.len() < res.len() {
-                res = window;
+            let sub_str = &s[start..end];
+            if min_str.len() == 0 || sub_str.len() < min_str.len() {
+                min_str = sub_str;
             }
 
             let start_letter = s.as_bytes()[start];
@@ -170,7 +153,7 @@ pub fn min_window(s: String, t: String) -> String {
         }
     }
 
-    res.to_string()
+    min_str.to_string()
 }
 ```
 

leetcode-docs/medium/12-int-to-roman.mdx

@@ -54,13 +54,13 @@ sidebar_label: 12. 整数转罗马数字
 ```ts
 输入: num = 58
 输出: 'LVIII'
-解释: L = 50, V = 5, III = 3.
+解释: (L = 50), (V = 5), (III = 3)
 ```
 
 ```ts
 输入: num = 1994
 输出: 'MCMXCIV'
-解释: M = 1000, CM = 900, XC = 90, IV = 4.
+解释: (M = 1000), (CM = 900), (XC = 90), (IV = 4)
 ```
 
 :::
@@ -81,7 +81,6 @@ import TabItem from '@theme/TabItem'
  * @return {string}
  */
 var intToRoman = function (num) {
-  let roman = ''
   const map = [
     [1000, 'M'],
     [900, 'CM'],
@@ -98,6 +97,7 @@ var intToRoman = function (num) {
     [1, 'I'],
   ]
 
+  let roman = ''
   for (const [value, symbol] of map) {
     while (num >= value) {
       num -= value

leetcode-docs/medium/3-length-of-longest-substring.mdx

@@ -88,17 +88,12 @@ var lengthOfLongestSubstring = function (s) {
   let max = 0
 
   while (end < n) {
-    const endChar = s[end++]
+    const endCh = s[end++]
+    map.set(endCh, map.has(endCh) ? map.get(endCh) + 1 : 1)
 
-    if (map.has(endChar)) {
-      map.set(endChar, map.get(endChar) + 1)
-    } else {
-      map.set(endChar, 1)
-    }
-
-    while (map.get(endChar) > 1) {
-      const startChar = s[start++]
-      map.set(startChar, map.get(startChar) - 1)
+    while (map.get(endCh) > 1) {
+      const startCh = s[start++]
+      map.set(startCh, map.get(startCh) - 1)
     }
 
     max = Math.max(max, end - start)
@@ -123,14 +118,14 @@ pub fn length_of_longest_substring(s: String) -> i32 {
     let mut end = 0;
 
     while end < s.len() {
-        let end_char = s_bytes[end];
+        let end_ch = s_bytes[end];
         end += 1;
-        map.entry(end_char).and_modify(|e| *e += 1).or_insert(1);
+        map.entry(end_ch).and_modify(|e| *e += 1).or_insert(1);
 
-        while map[&end_char] > 1 {
-            let start_char = s_bytes[start];
+        while map[&end_ch] > 1 {
+            let start_ch = s_bytes[start];
             start += 1;
-            map.entry(start_char).and_modify(|e| *e -= 1);
+            map.entry(start_ch).and_modify(|e| *e -= 1);
         }
 
         max = cmp::max(end - start, max);

leetcode-docs/medium/6-convert.mdx

@@ -74,11 +74,11 @@ import TabItem from '@theme/TabItem'
  * @return {string}
  */
 var convert = function (s, numRows) {
-  const n = s.length
-
   // 如果 numRows 比 2 小, 是没法变成 Z 型的, 直接返回 s 即可
   if (numRows < 2) return s
 
+  const n = s.length
+
   // 初始化一个数组, 数组长度为 numRows
   const res = Array(numRows).fill('')
 
@@ -90,7 +90,7 @@ var convert = function (s, numRows) {
     const carry = i % cycle
 
     // 计算索引, 如果余数小于 numRows, 那就用这个余数即可, 否则用循环数减去该余数
-    const idx = carry < numRows ? carry : cycle - carry;
+    const idx = carry < numRows ? carry : cycle - carry
 
     res[idx] += s[i]
   }

leetcode-docs/medium/7-reverse.mdx

@@ -75,6 +75,7 @@ pub fn reverse(x: i32) -> i32 {
         if n > i32::INT_MAX / 10 || n < i32::INT_MIN / 10 {
             return 0;
         }
+
         n = n * 10 + (x % 10);
         x /= 10;
     }

src/leetcode/rust/src/easy/question_14.rs

@@ -1,5 +1,7 @@
 #[allow(unused)]
 pub fn longest_common_prefix(strs: Vec<String>) -> String {
+    let mut strs = strs;
+    strs.sort();
     let mut prefix: &str = &strs[0];
     let mut has_found = false;
 

src/leetcode/rust/src/main.rs

@@ -3,7 +3,11 @@ mod hard;
 mod medium;
 
 fn main() {
-    let res = medium::question_40::combination_sum2(vec![10, 1, 2, 7, 6, 1, 5], 8);
+    let res = easy::question_14::longest_common_prefix(vec![
+        "flow".to_string(),
+        "flower".to_string(),
+        "flight".to_string(),
+    ]);
 
     println!("{:?}", res);
 }