第一题：双指针 void core(string &str) {     int len = str.size(), cur, right;     cur = right = len - 1;     for (; cur >= 0; cur--)     {         if (str[cur] == '#')             continue;         else         {             if (cur != right)                 swap(str[cur], str[right]);             right--;         }     } } 第二题：动态规划 略 第三题：最小公共祖先+dfs  代码未测试 struct TreeNode {     int val;          TreeNode *left;     TreeNode *right;     TreeNode(int x) : val(x), left(NULL), right(NULL) {} }; TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {     if (!root || root == p || root == q) return root;     TreeNode *left = lowestCommonAncestor(root->left, p, q);     TreeNode *right = lowestCommonAncestor(root->right, p, q);     if (!left && !right)         return root;     if (!right) return right;     else return left; } bool dfs(TreeNode *root, TreeNode *p, TreeNode *q, vector<TreeNode*> &path, vector<vector<TreeNode*>> &res) {     if (!root) return false;     if (root == p || root == q)     {         path.push_back(root);         res.push_back(path);         path.pop_back();         return true;     }     path.push_back(root);     if (dfs(root->left, p, q, path, res)) return true;     if (dfs(root->right, p, q, path, res)) return true;     path.pop_back();     return false; } vector<TreeNode*> findPath(TreeNode* root, TreeNode* p, TreeNode* q) {     TreeNode *ancestor = lowestCommonAncestor(root, p, q);     vector<TreeNode*> left_path, right_path;     vector<vector<TreeNode*>> res;     dfs(ancestor->left, p, q, left_path, res);     dfs(ancestor->right, p, q, right_path, res);     vector<TreeNode*> path;     for (int i = res[0].size() - 1; i >= 0; i--)         path.push_back(res[0][i]);     path.push_back(ancestor);     for (int i = 0; i < (int)res[1].size() - 1; i++)         path.push_back(res[1][i]);          return path; }