feat: update lc problems

solution/0100-0199/0161.One Edit Distance/README.md

@@ -50,7 +50,7 @@
 
 记 $m$ 表示字符串 $s$ 的长度，$n$ 表示字符串 $t$ 的长度。我们可以假定 $m$ 恒大于等于 $n$。
 
-若 $m-n\gt1$，直接返回 false；
+若 $m-n \gt 1$，直接返回 false；
 
 否则，遍历 $s$ 和 $t$，若遇到 $s[i]$ 不等于 $t[i]$：
 
@@ -59,7 +59,7 @@
 
 遍历结束，说明遍历过的 $s$ 跟 $t$ 所有字符相等，此时需要满足 $m=n+1$。
 
-时间复杂度 $O(m)$，空间复杂度 $O(1)$。
+时间复杂度 $O(m)$，其中 $m$ 为字符串 $s$ 的长度。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -139,6 +139,24 @@ func isOneEditDistance(s string, t string) bool {
 }
 ```
 
+```ts
+function isOneEditDistance(s: string, t: string): boolean {
+    const [m, n] = [s.length, t.length];
+    if (m < n) {
+        return isOneEditDistance(t, s);
+    }
+    if (m - n > 1) {
+        return false;
+    }
+    for (let i = 0; i < n; ++i) {
+        if (s[i] !== t[i]) {
+            return s.slice(i + 1) === t.slice(i + (m === n ? 1 : 0));
+        }
+    }
+    return m === n + 1;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- end -->

solution/0100-0199/0161.One Edit Distance/README_EN.md

@@ -43,7 +43,20 @@
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Discuss Different Cases
+
+Let $m$ represent the length of string $s$, and $n$ represent the length of string $t$. We can assume that $m$ is always greater than or equal to $n$.
+
+If $m-n > 1$, return false directly;
+
+Otherwise, iterate through $s$ and $t$, if $s[i]$ is not equal to $t[i]$:
+
+-   If $m \neq n$, compare $s[i+1:]$ with $t[i:]$, return true if they are equal, otherwise return false;
+-   If $m = n$, compare $s[i:]$ with $t[i:]$, return true if they are equal, otherwise return false.
+
+If the iteration ends, it means that all the characters of $s$ and $t$ that have been iterated are equal, at this time it needs to satisfy $m=n+1$.
+
+The time complexity is $O(m)$, where $m$ is the length of string $s$. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
@@ -123,6 +136,24 @@ func isOneEditDistance(s string, t string) bool {
 }
 ```
 
+```ts
+function isOneEditDistance(s: string, t: string): boolean {
+    const [m, n] = [s.length, t.length];
+    if (m < n) {
+        return isOneEditDistance(t, s);
+    }
+    if (m - n > 1) {
+        return false;
+    }
+    for (let i = 0; i < n; ++i) {
+        if (s[i] !== t[i]) {
+            return s.slice(i + 1) === t.slice(i + (m === n ? 1 : 0));
+        }
+    }
+    return m === n + 1;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- end -->

solution/0100-0199/0161.One Edit Distance/Solution.ts

@@ -0,0 +1,15 @@
+function isOneEditDistance(s: string, t: string): boolean {
+    const [m, n] = [s.length, t.length];
+    if (m < n) {
+        return isOneEditDistance(t, s);
+    }
+    if (m - n > 1) {
+        return false;
+    }
+    for (let i = 0; i < n; ++i) {
+        if (s[i] !== t[i]) {
+            return s.slice(i + 1) === t.slice(i + (m === n ? 1 : 0));
+        }
+    }
+    return m === n + 1;
+}

solution/0100-0199/0164.Maximum Gap/README.md

@@ -55,7 +55,7 @@ $$
 
 遍历数组结束之后，每个非空的桶内的最小值和最大值都可以确定。按照桶的编号从小到大的顺序依次遍历每个桶，当前的桶的最小值和上一个非空的桶的最大值是排序后的相邻元素，计算两个相邻元素之差，并更新最大间距。遍历桶结束之后即可得到最大间距。
 
-时间复杂度 $O(n)$，空间复杂度 $O(n)$。
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是数组 $nums$ 的长度。
 
 <!-- tabs:start -->
 

solution/0100-0199/0164.Maximum Gap/README_EN.md

@@ -37,7 +37,20 @@
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Discuss Different Cases
+
+Let $m$ represent the length of string $s$, and $n$ represent the length of string $t$. We can assume that $m$ is always greater than or equal to $n$.
+
+If $m-n > 1$, return false directly;
+
+Otherwise, iterate through $s$ and $t$, if $s[i]$ is not equal to $t[i]$:
+
+-   If $m \neq n$, compare $s[i+1:]$ with $t[i:]$, return true if they are equal, otherwise return false;
+-   If $m = n$, compare $s[i:]$ with $t[i:]$, return true if they are equal, otherwise return false.
+
+If the iteration ends, it means that all the characters of $s$ and $t$ that have been iterated are equal, at this time it needs to satisfy $m=n+1$.
+
+The time complexity is $O(m)$, where $m$ is the length of string $s$. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 

solution/0100-0199/0165.Compare Version Numbers/README.md

@@ -159,13 +159,16 @@ func compareVersion(version1 string, version2 string) int {
 
 ```ts
 function compareVersion(version1: string, version2: string): number {
-    let v1 = version1.split('.'),
-        v2 = version2.split('.');
-    for (let i = 0; i < Math.max(v1.length, v2.length); i++) {
-        let c1 = Number(v1[i] || 0),
-            c2 = Number(v2[i] || 0);
-        if (c1 > c2) return 1;
-        if (c1 < c2) return -1;
+    const v1 = version1.split('.');
+    const v2 = version2.split('.');
+    for (let i = 0; i < Math.max(v1.length, v2.length); ++i) {
+        const [n1, n2] = [+v1[i] || 0, +v2[i] || 0];
+        if (n1 < n2) {
+            return -1;
+        }
+        if (n1 > n2) {
+            return 1;
+        }
     }
     return 0;
 }

solution/0100-0199/0165.Compare Version Numbers/README_EN.md

@@ -152,13 +152,16 @@ func compareVersion(version1 string, version2 string) int {
 
 ```ts
 function compareVersion(version1: string, version2: string): number {
-    let v1 = version1.split('.'),
-        v2 = version2.split('.');
-    for (let i = 0; i < Math.max(v1.length, v2.length); i++) {
-        let c1 = Number(v1[i] || 0),
-            c2 = Number(v2[i] || 0);
-        if (c1 > c2) return 1;
-        if (c1 < c2) return -1;
+    const v1 = version1.split('.');
+    const v2 = version2.split('.');
+    for (let i = 0; i < Math.max(v1.length, v2.length); ++i) {
+        const [n1, n2] = [+v1[i] || 0, +v2[i] || 0];
+        if (n1 < n2) {
+            return -1;
+        }
+        if (n1 > n2) {
+            return 1;
+        }
     }
     return 0;
 }

solution/0100-0199/0165.Compare Version Numbers/Solution.ts

@@ -1,11 +1,14 @@
 function compareVersion(version1: string, version2: string): number {
-    let v1 = version1.split('.'),
-        v2 = version2.split('.');
-    for (let i = 0; i < Math.max(v1.length, v2.length); i++) {
-        let c1 = Number(v1[i] || 0),
-            c2 = Number(v2[i] || 0);
-        if (c1 > c2) return 1;
-        if (c1 < c2) return -1;
+    const v1 = version1.split('.');
+    const v2 = version2.split('.');
+    for (let i = 0; i < Math.max(v1.length, v2.length); ++i) {
+        const [n1, n2] = [+v1[i] || 0, +v2[i] || 0];
+        if (n1 < n2) {
+            return -1;
+        }
+        if (n1 > n2) {
+            return 1;
+        }
     }
     return 0;
 }