Merge branch 'master' into master

tlylt · May 28, 2022 · 2b434d0 · 2b434d0
2 parents 35666ef + f6456c3
commit 2b434d0
Show file tree

Hide file tree

Showing 22 changed files with 747 additions and 11 deletions.
diff --git a/problems/0019.删除链表的倒数第N个节点.md b/problems/0019.删除链表的倒数第N个节点.md
@@ -39,7 +39,7 @@
 
 分为如下几步：
 
-* 首先这里我推荐大家使用虚拟头结点，这样方面处理删除实际头结点的逻辑，如果虚拟头结点不清楚，可以看这篇： [链表：听说用虚拟头节点会方便很多？](https://programmercarl.com/0203.移除链表元素.html)
+* 首先这里我推荐大家使用虚拟头结点，这样方便处理删除实际头结点的逻辑，如果虚拟头结点不清楚，可以看这篇： [链表：听说用虚拟头节点会方便很多？](https://programmercarl.com/0203.移除链表元素.html)
 
 * 定义fast指针和slow指针，初始值为虚拟头结点，如图：
 

diff --git a/problems/0027.移除元素.md b/problems/0027.移除元素.md
@@ -81,7 +81,7 @@ public:
 
 **双指针法（快慢指针法）在数组和链表的操作中是非常常见的，很多考察数组、链表、字符串等操作的面试题，都使用双指针法。**
 
-后序都会一一介绍到，本题代码如下：
+后续都会一一介绍到，本题代码如下：
 
 ```CPP
 // 时间复杂度：O(n)

diff --git a/problems/0059.螺旋矩阵II.md b/problems/0059.螺旋矩阵II.md
@@ -564,6 +564,57 @@ int** generateMatrix(int n, int* returnSize, int** returnColumnSizes){
     return ans;
 }
 ```
-
+Scala:
+```scala
+object Solution {
+  def generateMatrix(n: Int): Array[Array[Int]] = {
+    var res = Array.ofDim[Int](n, n) // 定义一个n*n的二维矩阵
+    var num = 1 // 标志当前到了哪个数字
+    var i = 0 // 横坐标
+    var j = 0 // 竖坐标
+
+    while (num <= n * n) {
+      // 向右：当j不越界，并且下一个要填的数字是空白时
+      while (j < n && res(i)(j) == 0) {
+        res(i)(j) = num // 当前坐标等于num
+        num += 1 // num++
+        j += 1 // 竖坐标+1
+      }
+      i += 1 // 下移一行
+      j -= 1 // 左移一列
+
+      // 剩下的都同上
+
+      // 向下
+      while (i < n && res(i)(j) == 0) {
+        res(i)(j) = num
+        num += 1
+        i += 1
+      }
+      i -= 1
+      j -= 1
+
+      // 向左
+      while (j >= 0 && res(i)(j) == 0) {
+        res(i)(j) = num
+        num += 1
+        j -= 1
+      }
+      i -= 1
+      j += 1
+
+      // 向上
+      while (i >= 0 && res(i)(j) == 0) {
+        res(i)(j) = num
+        num += 1
+        i -= 1
+      }
+      i += 1
+      j += 1
+    }
+    res
+  }
+}
+```
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
diff --git a/problems/0070.爬楼梯完全背包版本.md b/problems/0070.爬楼梯完全背包版本.md
@@ -199,6 +199,28 @@ var climbStairs = function(n) {
 };
 ```
 
+TypeScript：
+
+```typescript
+function climbStairs(n: number): number {
+    const m: number = 2;    // 本题m为2
+    const dp: number[] = new Array(n + 1).fill(0);
+    dp[0] = 1;
+    // 遍历背包
+    for (let i = 1; i <= n; i++) {
+        // 遍历物品
+        for (let j = 1; j <= m; j++) {
+            if (j <= i) {
+                dp[i] += dp[i - j];
+            }
+        }
+    }
+    return dp[n];
+};
+```
+
+
+
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md
@@ -82,6 +82,26 @@ public:
     }
 };
 ```
+```CPP
+# 递归法
+class Solution {
+public:
+    void order(TreeNode* cur, vector<vector<int>>& result, int depth)
+    {
+        if (cur == nullptr) return;
+        if (result.size() == depth) result.push_back(vector<int>());
+        result[depth].push_back(cur->val);
+        order(cur->left, result, depth + 1);
+        order(cur->right, result, depth + 1);
+    }
+    vector<vector<int>> levelOrder(TreeNode* root) {
+        vector<vector<int>> result;
+        int depth = 0;
+        order(root, result, depth);
+        return result;
+    }
+};
+```
 
 python3代码：
 

diff --git a/problems/0112.路径总和.md b/problems/0112.路径总和.md
@@ -377,22 +377,22 @@ class solution {
 
 ```java
 class solution {
-    public list<list<integer>> pathsum(treenode root, int targetsum) {
-        list<list<integer>> res = new arraylist<>();
+    public List<List<Integer>> pathsum(TreeNode root, int targetsum) {
+        List<List<Integer>> res = new ArrayList<>();
         if (root == null) return res; // 非空判断
-        
-        list<integer> path = new linkedlist<>();
+
+        List<Integer> path = new LinkedList<>();
         preorderdfs(root, targetsum, res, path);
         return res;
     }
 
-    public void preorderdfs(treenode root, int targetsum, list<list<integer>> res, list<integer> path) {
+    public void preorderdfs(TreeNode root, int targetsum, List<List<Integer>> res, List<Integer> path) {
         path.add(root.val);
         // 遇到了叶子节点
         if (root.left == null && root.right == null) {
             // 找到了和为 targetsum 的路径
             if (targetsum - root.val == 0) {
-                res.add(new arraylist<>(path));
+                res.add(new ArrayList<>(path));
             }
             return; // 如果和不为 targetsum，返回
         }
@@ -1006,6 +1006,126 @@ func traversal(_ cur: TreeNode?, count: Int) {
 }
 ```
 
+## C
+> 0112.路径总和
+递归法：
+```c
+bool hasPathSum(struct TreeNode* root, int targetSum){
+    // 递归结束条件：若当前节点不存在，返回false
+    if(!root)
+        return false;
+    // 若当前节点为叶子节点，且targetSum-root的值为0。（当前路径上的节点值的和满足条件）返回true
+    if(!root->right && !root->left && targetSum == root->val)
+        return true;
+
+    // 查看左子树和右子树的所有节点是否满足条件
+    return hasPathSum(root->right, targetSum - root->val) || hasPathSum(root->left, targetSum - root->val);
+}
+```
+
+迭代法:
+```c
+// 存储一个节点以及当前的和
+struct Pair {
+    struct TreeNode* node;
+    int sum;
+};
+
+bool hasPathSum(struct TreeNode* root, int targetSum){
+    struct Pair stack[1000];
+    int stackTop = 0;
+
+    // 若root存在，则将节点和值封装成一个pair入栈
+    if(root) {
+        struct Pair newPair = {root, root->val};
+        stack[stackTop++] = newPair;
+    }
+
+    // 当栈不为空时
+    while(stackTop) {
+        // 出栈栈顶元素
+        struct Pair topPair = stack[--stackTop];
+        // 若栈顶元素为叶子节点，且和为targetSum时，返回true
+        if(!topPair.node->left && !topPair.node->right && topPair.sum == targetSum)
+            return true;
+        
+        // 若当前栈顶节点有左右孩子，计算和并入栈
+        if(topPair.node->left) {
+            struct Pair newPair = {topPair.node->left, topPair.sum + topPair.node->left->val};
+            stack[stackTop++] = newPair;
+        }
+        if(topPair.node->right) {
+            struct Pair newPair = {topPair.node->right, topPair.sum + topPair.node->right->val};
+            stack[stackTop++] = newPair;
+        }
+    }
+    return false;
+}
+```
+> 0113.路径总和 II
+```c
+int** ret;
+int* path;
+int* colSize;
+int retTop;
+int pathTop;
+
+void traversal(const struct TreeNode* const node, int count) {
+    // 若当前节点为叶子节点
+    if(!node->right && !node->left) {
+        // 若当前path上的节点值总和等于targetSum。
+        if(count == 0) {
+            // 复制当前path
+            int *curPath = (int*)malloc(sizeof(int) * pathTop);
+            memcpy(curPath, path, sizeof(int) * pathTop);
+            // 记录当前path的长度为pathTop
+            colSize[retTop] = pathTop;
+            // 将当前path加入到ret数组中
+            ret[retTop++] = curPath;
+        }
+        return;
+    }
+
+    // 若节点有左/右孩子
+    if(node->left) {
+        // 将左孩子的值加入path中
+        path[pathTop++] = node->left->val;
+        traversal(node->left, count - node->left->val);
+        // 回溯
+        pathTop--;
+    }
+    if(node->right) {
+        // 将右孩子的值加入path中
+        path[pathTop++] = node->right->val;
+        traversal(node->right, count - node->right->val);
+        // 回溯
+        --pathTop;
+    }
+}
+
+int** pathSum(struct TreeNode* root, int targetSum, int* returnSize, int** returnColumnSizes){
+    // 初始化数组
+    ret = (int**)malloc(sizeof(int*) * 1000);
+    path = (int*)malloc(sizeof(int*) * 1000);
+    colSize = (int*)malloc(sizeof(int) * 1000);
+    retTop = pathTop = 0;
+    *returnSize = 0;
+
+    // 若根节点不存在，返回空的ret
+    if(!root)
+        return ret;
+    // 将根节点加入到path中
+    path[pathTop++] = root->val;
+    traversal(root, targetSum - root->val);
+
+    // 设置返回ret数组大小，以及其中每个一维数组元素的长度
+    *returnSize = retTop;
+    *returnColumnSizes = colSize;
+
+    return ret;
+}
+```
+
 
 
 -----------------------

diff --git a/problems/0139.单词拆分.md b/problems/0139.单词拆分.md
@@ -345,6 +345,48 @@ const wordBreak = (s, wordDict) => {
 }
 ```
 
+TypeScript：
+
+> 动态规划
+
+```typescript
+function wordBreak(s: string, wordDict: string[]): boolean {
+    const dp: boolean[] = new Array(s.length + 1).fill(false);
+    dp[0] = true;
+    for (let i = 1; i <= s.length; i++) {
+        for (let j = 0; j < i; j++) {
+            const tempStr: string = s.slice(j, i);
+            if (wordDict.includes(tempStr) && dp[j] === true) {
+                dp[i] = true;
+                break;
+            }
+        }
+    }
+    return dp[s.length];
+};
+```
+
+> 记忆化回溯
+
+```typescript
+function wordBreak(s: string, wordDict: string[]): boolean {
+    // 只需要记忆结果为false的情况
+    const memory: boolean[] = [];
+    return backTracking(s, wordDict, 0, memory);
+    function backTracking(s: string, wordDict: string[], startIndex: number, memory: boolean[]): boolean {
+        if (startIndex >= s.length) return true;
+        if (memory[startIndex] === false) return false;
+        for (let i = startIndex + 1, length = s.length; i <= length; i++) {
+            const str: string = s.slice(startIndex, i);
+            if (wordDict.includes(str) && backTracking(s, wordDict, i, memory))
+                return true;
+        }
+        memory[startIndex] = false;
+        return false;
+    }
+};
+```
+
 
 
 -----------------------

diff --git a/problems/0198.打家劫舍.md b/problems/0198.打家劫舍.md
@@ -189,6 +189,29 @@ const rob = nums => {
 };
 ```
 
+TypeScript：
+
+```typescript
+function rob(nums: number[]): number {
+    /**
+        dp[i]: 前i个房屋能偷到的最大金额
+        dp[0]: nums[0];
+        dp[1]: max(nums[0], nums[1]);
+        ...
+        dp[i]: max(dp[i-1], dp[i-2]+nums[i]);
+     */
+    const length: number = nums.length;
+    if (length === 1) return nums[0];
+    const dp: number[] = [];
+    dp[0] = nums[0];
+    dp[1] = Math.max(nums[0], nums[1]);
+    for (let i = 2; i < length; i++) {
+        dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]);
+    }
+    return dp[length - 1];
+};
+```
+