[pull] master from youngyangyang04:master

problems/0459.重复的子字符串.md

@@ -689,6 +689,29 @@ var repeatedSubstringPattern = function (s) {
 };
 ```
 
+> 正则匹配
+```javascript
+/**
+ * @param {string} s
+ * @return {boolean}
+ */
+var repeatedSubstringPattern = function(s) {
+  let reg = /^(\w+)\1+$/
+  return reg.test(s)
+};
+```
+> 移动匹配
+```javascript
+/**
+ * @param {string} s
+ * @return {boolean}
+ */
+var repeatedSubstringPattern = function (s) {
+    let ss = s + s;
+    return ss.substring(1, ss.length - 1).includes(s);
+};
+```
+
 ### TypeScript:
 
 > 前缀表统一减一
@@ -894,8 +917,10 @@ impl Solution {
 }
 ```
 ### C#
+
+> 前缀表不减一
+
 ```csharp
-// 前缀表不减一
 public bool RepeatedSubstringPattern(string s)
 {
     if (s.Length == 0)
@@ -920,6 +945,13 @@ public int[] GetNext(string s)
 }
 ```
 
+> 移动匹配
+```csharp
+public bool RepeatedSubstringPattern(string s) {
+    string  ss = (s + s).Substring(1, (s + s).Length - 2);
+    return ss.Contains(s);
+}
+```
 ### C
 
 ```c

problems/背包问题理论基础完全背包.md

@@ -316,7 +316,51 @@ print(knapsack(n, bag_weight, weight, value))
 
 ### JavaScript
 
+```js
+const readline = require('readline').createInterface({
+    input: process.stdin,
+    output: process.stdout
+});
+
+let input = [];
+readline.on('line', (line) => {
+    input.push(line.trim());
+});
+
+readline.on('close', () => {
+    // 第一行解析 n 和 v
+    const [n, bagweight] = input[0].split(' ').map(Number);
+
+    /// 剩余 n 行解析重量和价值
+    const weight = [];
+    const value = [];
+    for (let i = 1; i <= n; i++) {
+        const [wi, vi] = input[i].split(' ').map(Number);
+        weight.push(wi);
+        value.push(vi);
+    }
+
+
+    let dp = Array.from({ length: n }, () => Array(bagweight + 1).fill(0));
+
+    for (let j = weight[0]; j <= bagweight; j++) {
+        dp[0][j] = dp[0][j-weight[0]] + value[0];
+    }
+
+    for (let i = 1; i < n; i++) {
+        for (let j = 0; j <= bagweight; j++) {
+            if (j < weight[i]) {
+                dp[i][j] = dp[i - 1][j];
+            } else {
+                dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - weight[i]] + value[i]);
+            }
+        }
+    }
 
+    console.log(dp[n - 1][bagweight]);
+});
+
+```
 
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