Merge branch 'youngyangyang04:master' into master

problems/0300.最长上升子序列.md

@@ -168,6 +168,39 @@ func lengthOfLIS(nums []int ) int {
 }
 ```
 
+```go
+// 动态规划求解
+func lengthOfLIS(nums []int) int {
+    // dp数组的定义 dp[i]表示取第i个元素的时候，表示子序列的长度，其中包括 nums[i] 这个元素
+    dp := make([]int, len(nums))
+
+    // 初始化，所有的元素都应该初始化为1
+    for i := range dp {
+        dp[i] = 1
+    }
+
+    ans := dp[0]
+    for i := 1; i < len(nums); i++ {
+        for j := 0; j < i; j++ {
+            if nums[i] > nums[j] {
+                dp[i] = max(dp[i], dp[j] + 1)
+            }
+        }
+        if dp[i] > ans {
+            ans = dp[i]
+        }
+    }
+    return ans
+}
+
+func max(x, y int) int {
+    if x > y {
+        return x
+    }
+    return y
+}
+```
+
 Javascript
 ```javascript
 const lengthOfLIS = (nums) => {

problems/0674.最长连续递增序列.md

@@ -236,6 +236,45 @@ class Solution:
 ```
 
 Go：
+> 动态规划：
+```go
+func findLengthOfLCIS(nums []int) int {
+	if len(nums) == 0 {return 0}
+	res, count := 1, 1
+	for i := 0; i < len(nums)-1; i++ {
+		if nums[i+1] > nums[i] {
+			count++
+		}else {
+			count = 1
+		}
+		if count > res {
+			res = count
+		}
+	}
+	return res
+}
+```
+
+> 贪心算法：
+```go
+func findLengthOfLCIS(nums []int) int {
+	if len(nums) == 0 {return 0}
+	dp := make([]int, len(nums))
+	for i := 0; i < len(dp); i++ {
+		dp[i] = 1
+	}
+	res := 1
+	for i := 0; i < len(nums)-1; i++ {
+		if nums[i+1] > nums[i] {
+			dp[i+1] = dp[i] + 1
+		}
+		if dp[i+1] > res {
+			res = dp[i+1]
+		}
+	}
+	return res
+}
+```
 
 Javascript：
 

problems/0968.监控二叉树.md

@@ -476,7 +476,35 @@ var minCameraCover = function(root) {
 };
 ```
 
+### TypeScript
+
+```typescript
+function minCameraCover(root: TreeNode | null): number {
+    /** 0-无覆盖， 1-有摄像头， 2-有覆盖 */
+    type statusCode = 0 | 1 | 2;
+    let resCount: number = 0;
+    if (recur(root) === 0) resCount++;
+    return resCount;
+    function recur(node: TreeNode | null): statusCode {
+        if (node === null) return 2;
+        const left: statusCode = recur(node.left),
+            right: statusCode = recur(node.right);
+        let resStatus: statusCode = 0;
+        if (left === 0 || right === 0) {
+            resStatus = 1;
+            resCount++;
+        } else if (left === 1 || right === 1) {
+            resStatus = 2;
+        } else {
+            resStatus = 0;
+        }
+        return resStatus;
+    }
+};
+```
+
 ### C 
+
 ```c
 /*
 **函数后序遍历二叉树。判断一个结点状态时，根据其左右孩子结点的状态进行判断

problems/二叉树的迭代遍历.md

@@ -11,9 +11,9 @@
 
 看完本篇大家可以使用迭代法，再重新解决如下三道leetcode上的题目：
 
-* 144.二叉树的前序遍历
+* [144.二叉树的前序遍历](https://leetcode-cn.com/problems/binary-tree-preorder-traversal/)
-* 94.二叉树的中序遍历
+* [94.二叉树的中序遍历](https://leetcode-cn.com/problems/binary-tree-inorder-traversal/)
-* 145.二叉树的后序遍历
+* [145.二叉树的后序遍历](https://leetcode-cn.com/problems/binary-tree-postorder-traversal/)
 
 为什么可以用迭代法（非递归的方式）来实现二叉树的前后中序遍历呢？
 

problems/二叉树的递归遍历.md

@@ -99,9 +99,9 @@ void traversal(TreeNode* cur, vector<int>& vec) {
 
 此时大家可以做一做leetcode上三道题目，分别是：
 
-* 144.二叉树的前序遍历
+* [144.二叉树的前序遍历](https://leetcode-cn.com/problems/binary-tree-preorder-traversal/)
-* 145.二叉树的后序遍历
+* [145.二叉树的后序遍历](https://leetcode-cn.com/problems/binary-tree-postorder-traversal/)
-* 94.二叉树的中序遍历
+* [94.二叉树的中序遍历](https://leetcode-cn.com/problems/binary-tree-inorder-traversal/)
 
 可能有同学感觉前后中序遍历的递归太简单了，要打迭代法（非递归），别急，我们明天打迭代法，打个通透！
 

problems/背包理论基础01背包-1.md

@@ -380,28 +380,37 @@ func main() {
 ### javascript 
 
 ```js
-function testweightbagproblem (wight, value, size) {
+/**
-  const len = wight.length, 
+ * 
-    dp = array.from({length: len + 1}).map(
+ * @param {Number []} weight 
-      () => array(size + 1).fill(0)
+ * @param {Number []} value 
-    );
+ * @param {Number} size 
+ * @returns 
+ */
 
-  for(let i = 1; i <= len; i++) {
+function testWeightBagProblem(weight, value, size) {
-    for(let j = 0; j <= size; j++) {
+const len = weight.length,
-      if(wight[i - 1] <= j) {
+dp = Array.from({length: len}).map(
-        dp[i][j] = math.max(
+() => Array(size + 1))  //JavaScript 数组是引用类型	
-          dp[i - 1][j], 
+for(let i = 0; i < len; i++) {  //初始化最左一列，即背包容量为0时的情况
-          value[i - 1] + dp[i - 1][j - wight[i - 1]]
+dp[i][0] = 0;
-        )
+}
-      } else {
+for(let j = 1; j < size+1; j++) { //初始化第0行, 只有一件物品的情况
-        dp[i][j] = dp[i - 1][j];
+if(weight[0] <= j) {
-      }
+dp[0][j] = value[0];
-    }
+} else {
-  }
+dp[0][j] = 0;
+}
+}
 	
-//   console.table(dp);
+for(let i = 1; i < len; i++) {  //dp[i][j]由其左上方元素推导得出
+for(let j = 1; j < size+1; j++) {
+if(j < weight[i]) dp[i][j] = dp[i - 1][j];
+else dp[i][j] = Math.max(dp[i-1][j], dp[i-1][j - weight[i]] + value[i]);
+}
+}
 
-  return dp[len][size];
+return dp[len-1][size] //满足条件的最大值
 }
 
 function testWeightBagProblem2 (wight, value, size) {