feat: update lc problems

Author: yanglbme

solution/0100-0199/0179.Largest Number/README.md

@@ -128,7 +128,7 @@ public class Comparer: IComparer<string>
     {
         return Compare(left, right, 0, 0);
     }
-    
+
     private int Compare(string left, string right, int lBegin, int rBegin)
     {
         var len = Math.Min(left.Length - lBegin, right.Length - rBegin);
@@ -139,7 +139,7 @@ public class Comparer: IComparer<string>
                 return left[lBegin + i] < right[rBegin + i] ? -1 : 1;
             }
         }
-        
+
         if (left.Length - lBegin == right.Length - rBegin)
         {
             return 0;
@@ -159,7 +159,7 @@ public class Solution {
     public string LargestNumber(int[] nums) {
         var sb = new StringBuilder();
         var strs = nums.Select(n => n.ToString(CultureInfo.InvariantCulture)).OrderByDescending(s => s, new Comparer());
-        
+
         var nonZeroOccurred = false;
         foreach (var str in strs)
         {

solution/0100-0199/0179.Largest Number/README_EN.md

@@ -115,7 +115,7 @@ public class Comparer: IComparer<string>
     {
         return Compare(left, right, 0, 0);
     }
-    
+
     private int Compare(string left, string right, int lBegin, int rBegin)
     {
         var len = Math.Min(left.Length - lBegin, right.Length - rBegin);
@@ -126,7 +126,7 @@ public class Comparer: IComparer<string>
                 return left[lBegin + i] < right[rBegin + i] ? -1 : 1;
             }
         }
-        
+
         if (left.Length - lBegin == right.Length - rBegin)
         {
             return 0;
@@ -146,7 +146,7 @@ public class Solution {
     public string LargestNumber(int[] nums) {
         var sb = new StringBuilder();
         var strs = nums.Select(n => n.ToString(CultureInfo.InvariantCulture)).OrderByDescending(s => s, new Comparer());
-        
+
         var nonZeroOccurred = false;
         foreach (var str in strs)
         {

solution/0300-0399/0315.Count of Smaller Numbers After Self/README_EN.md

@@ -4,7 +4,7 @@
 
 ## Description
 
-<p>You are given an integer array <code>nums</code> and you have to return a new <code>counts</code> array. The <code>counts</code> array has the property where <code>counts[i]</code> is the number of smaller elements to the right of <code>nums[i]</code>.</p>
+<p>Given an integer array <code>nums</code>, return<em> an integer array </em><code>counts</code><em> where </em><code>counts[i]</code><em> is the number of smaller elements to the right of </em><code>nums[i]</code>.</p>
 
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>

solution/0300-0399/0347.Top K Frequent Elements/README_EN.md

@@ -19,6 +19,7 @@
 
 <ul>
 	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
 	<li><code>k</code> is in the range <code>[1, the number of unique elements in the array]</code>.</li>
 	<li>It is <strong>guaranteed</strong> that the answer is <strong>unique</strong>.</li>
 </ul>

solution/0500-0599/0593.Valid Square/README.md

@@ -8,7 +8,7 @@
 
 <p>给定2D空间中四个点的坐标&nbsp;<code>p1</code>,&nbsp;<code>p2</code>,&nbsp;<code>p3</code>&nbsp;和&nbsp;<code>p4</code>，如果这四个点构成一个正方形，则返回 <code>true</code> 。</p>
 
-<p>点的坐标&nbsp;<code>p<sub>i</sub></code> 表示为 <code>[xi, yi]</code> 。输入 <strong>不是</strong> 按任何顺序给出的。</p>
+<p>点的坐标&nbsp;<code>p<sub>i</sub></code> 表示为 <code>[xi, yi]</code> 。 <code>输入没有任何顺序</code> 。</p>
 
 <p>一个 <strong>有效的正方形</strong> 有四条等边和四个等角(90度角)。</p>
 

solution/0600-0699/0622.Design Circular Queue/README.md

@@ -195,37 +195,37 @@ public:
         q = vector<int>(k);
         front = size = 0;
     }
-    
+
     bool enQueue(int value) {
         if (isFull()) return false;
         int idx = (front + size) % capacity;
         q[idx] = value;
         ++size;
         return true;
     }
-    
+
     bool deQueue() {
         if (isEmpty()) return false;
         front = (front + 1) % capacity;
         --size;
         return true;
     }
-    
+
     int Front() {
         if (isEmpty()) return -1;
         return q[front];
     }
-    
+
     int Rear() {
         if (isEmpty()) return -1;
         int idx = (front + size - 1) % capacity;
         return q[idx];
     }
-    
+
     bool isEmpty() {
         return size == 0;
     }
-    
+
     bool isFull() {
         return size == capacity;
     }

solution/0600-0699/0622.Design Circular Queue/README_EN.md

@@ -194,37 +194,37 @@ public:
         q = vector<int>(k);
         front = size = 0;
     }
-    
+
     bool enQueue(int value) {
         if (isFull()) return false;
         int idx = (front + size) % capacity;
         q[idx] = value;
         ++size;
         return true;
     }
-    
+
     bool deQueue() {
         if (isEmpty()) return false;
         front = (front + 1) % capacity;
         --size;
         return true;
     }
-    
+
     int Front() {
         if (isEmpty()) return -1;
         return q[front];
     }
-    
+
     int Rear() {
         if (isEmpty()) return -1;
         int idx = (front + size - 1) % capacity;
         return q[idx];
     }
-    
+
     bool isEmpty() {
         return size == 0;
     }
-    
+
     bool isFull() {
         return size == capacity;
     }

solution/0600-0699/0638.Shopping Offers/README_EN.md

@@ -39,11 +39,9 @@ You cannot add more items, though only $9 for 2A ,2B and 1C.
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li><code>n == price.length</code></li>
-	<li><code>n == needs.length</code></li>
+	<li><code>n == price.length == needs.length</code></li>
 	<li><code>1 &lt;= n &lt;= 6</code></li>
-	<li><code>0 &lt;= price[i] &lt;= 10</code></li>
-	<li><code>0 &lt;= needs[i] &lt;= 10</code></li>
+	<li><code>0 &lt;= price[i], needs[i] &lt;= 10</code></li>
 	<li><code>1 &lt;= special.length &lt;= 100</code></li>
 	<li><code>special[i].length == n + 1</code></li>
 	<li><code>0 &lt;= special[i][j] &lt;= 50</code></li>

solution/0800-0899/0802.Find Eventual Safe States/README_EN.md

@@ -322,7 +322,7 @@ var eventualSafeNodes = function (graph) {
  * @param {number[][]} graph
  * @return {number[]}
  */
- var eventualSafeNodes = function(graph) {
+var eventualSafeNodes = function (graph) {
     const n = graph.length;
     const color = new Array(n).fill(0);
     function dfs(i) {

solution/0800-0899/0802.Find Eventual Safe States/Solution.js

@@ -2,7 +2,7 @@
  * @param {number[][]} graph
  * @return {number[]}
  */
- var eventualSafeNodes = function(graph) {
+var eventualSafeNodes = function (graph) {
     const n = graph.length;
     const color = new Array(n).fill(0);
     function dfs(i) {
@@ -25,4 +25,4 @@
         }
     }
     return ans;
-};
+};