补交homework

Week_02/id_145/LeetCode_236_145.cpp

@@ -0,0 +1,27 @@
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
+ * };
+ */
+class Solution {
+public:
+    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
+        /**
+        注意p,q必然存在树内, 且所有节点的值唯一!!!
+        递归思想, 对以root为根的(子)树进行查找p和q, 如果root == null || p || q 直接返回root
+        表示对于当前树的查找已经完毕, 否则对左右子树进行查找, 根据左右子树的返回值判断:
+        1. 左右子树的返回值都不为null, 由于值唯一左右子树的返回值就是p和q, 此时root为LCA
+        2. 如果左右子树返回值只有一个不为null, 说明只有p和q存在与左或右子树中, 最先找到的那个节点为LCA
+        3. 左右子树返回值均为null, p和q均不在树中, 返回null
+        **/
+        if (!root || root == p || root == q) return root;
+        auto left = lowestCommonAncestor(root->left, p, q);
+        auto right = lowestCommonAncestor(root->right, p, q);
+        if (left && right) return root;
+        return left ? left : right;
+    }
+};

Week_02/id_145/LeetCode_692_145.cpp

@@ -0,0 +1,27 @@
+struct intComp {
+	bool operator() (const pair<int, string>& lhs, const pair<int, string>& rhs) const{
+		return lhs.first > rhs.first;
+	}
+};
+class Solution {
+public:
+    vector<string> topKFrequent(vector<string>& words, int k) {
+        vector<string> resVec(k);//存储结果
+		map<string, int> hashMap;
+        //用于统计各个单词出现的次数，并按照字母顺序排序
+		for (auto word : words) {
+			hashMap[word] += 1;
+		}
+        //然后按照单词出现的次数进行排序
+		multiset<pair<int, string>, intComp> mySet;
+		for (auto &item : hashMap) {
+			mySet.insert({ item.second, item.first });
+		}
+        //取出前k个
+		multiset<pair<int, string>>::iterator it = mySet.begin();
+		for (int i = 0; i < k; ++i, ++it) {
+			resVec[i] = it->second;
+		}
+		return resVec;
+    }
+};

Week_03/id_145/LeetCode_200_145.cpp

@@ -0,0 +1,31 @@
+class Solution {
+public:
+    int i,j,m,n,index=2;//index用于标记不同岛序号
+    const int left = 0;const int right = 1; //方向常量设置
+    const int up = 2;const int down = 3;
+    const int none = 4;
+    vector<vector<int>> map;
+    vector<int> tmp;
+    int numIslands(vector<vector<char>>& grid) {
+        if(!grid.size()) return 0;
+        m = grid.size();n = grid[0].size();
+        for(i=0;i<m;i++) {
+            tmp.clear();
+            for(j=0;j<n;j++) tmp.push_back(grid[i][j]-'0');
+            map.push_back(tmp);
+        }
+        for(i=0;i<m;i++) 
+           for(j=0;j<n;j++) 
+                if(map[i][j]==1) dfs(i,j,none),index++;
+        return index-2;
+    }
+    void dfs(int i,int j,int flag) { //flag判断来的方向，预防死环
+        if(i<0||i>m-1||j<0||j>n-1) return;
+        if(map[i][j]!=1) return;
+        map[i][j] = index;
+        if(flag!=right) dfs(i,j+1,left); 
+        if(flag!=left) dfs(i,j-1,right);
+        if(flag!=up) dfs(i+1,j,down);
+        if(flag!=down) dfs(i-1,j,up);
+    }
+};

Week_03/id_145/LeetCode_310_145.cpp

@@ -0,0 +1,33 @@
+class Solution {
+public:
+    vector<int> findMinHeightTrees(int n, vector<vector<int>>& edges) {
+        if(n==1) return {0};
+        vector<unordered_set<int> > adj(n);
+        vector<int> res;
+        queue<int> q;
+        for(auto edge:edges){
+            adj[edge[0]].insert(edge[1]);
+            adj[edge[1]].insert(edge[0]);
+        }
+        for(int i=0;i<n;i++){
+            if(adj[i].size()==1) q.push(i);
+        }
+        while(n>2){
+            int size=q.size();
+            n-=size;
+            while(size-->0){
+                auto t=q.front();
+                q.pop();
+                for(auto itm:adj[t]){
+                    adj[itm].erase(t);
+                    if(adj[itm].size()==1) q.push(itm);
+                }
+            }
+        }
+        while(!q.empty()){
+            res.push_back(q.front());
+            q.pop();
+        }
+        return res;
+    }
+};

Week_04/id_145/LeetCode_51_145.cpp

@@ -0,0 +1,58 @@
+class Solution {
+public:
+    vector<vector<string>> solveNQueens(int n) {
+        vector<vector<string>> res;
+        vector<vector<int>> mark;
+        vector<string> item;
+        for(int i = 0; i < n; i ++)
+        {
+            mark.push_back(vector<int>());
+            for(int j = 0; j < n; j ++)
+            {
+                mark[i].push_back(0);
+            }
+            item.push_back("");
+            item[i].append(n,'.');//添加n个'.'
+        }
+        generate(0, n, item, res, mark);
+        return res;
+    }
+    //回溯法
+    void generate(int k, int n, vector<string> &item, vector<vector<string>> &res, vector<vector<int>> &mark)
+    {
+        if(k == n)
+        {
+            res.push_back(item);
+            return;
+        }
+        for(int i = 0; i < n; i ++)
+        {
+            if(mark[k][i] == 0)
+            {
+                vector<vector<int>>  tmp_mark = mark;
+                item[k][i] = 'Q';
+                put_down_the_queen(k, i, mark);//更新mark矩阵
+                generate(k + 1, n, item, res, mark);
+                mark = tmp_mark;
+                item[k][i] = '.';
+            }
+        }
+    }
+    //放下queen后更新mark矩阵
+    void put_down_the_queen(int x, int y, vector<vector<int>> &mark)
+    {
+        static const int dx[] = {-1, 0, 1, -1, 1, -1, 0, 1};
+        static const int dy[] = {1, 1, 1, 0, 0, -1, -1, -1};
+        mark[x][y] = 1;
+        for(int i = 1; i < mark.size(); i ++)
+        {
+            for(int j = 0; j < 8; j ++)
+            {
+                int new_x = x + i * dx[j];
+                int new_y = y + i * dy[j];
+                if(new_x >= 0 && new_x < mark.size() && new_y >= 0 && new_y < mark.size()) 
+                    mark[new_x][new_y] = 1;
+            }
+        }
+    }
+};

Week_04/id_145/LeetCode_72_145.cpp

@@ -0,0 +1,26 @@
+class Solution {
+public:
+    int minDistance(string word1, string word2) {
+        int len1=word1.length(),len2=word2.length();
+        //dp[index1][index2]:前index1个word1字符转换成前index2个word2字符需要的最少步数
+        vector<vector<int>> dp(len1+1,vector<int>(len2+1));
+        for(int index1=0;index1<=len1;index1++){
+            for(int index2=0;index2<=len2;index2++){
+                if(index1 == 0)
+                    //从无到有显然要经历index2步插入操作
+                    dp[index1][index2] = index2;
+                else if(index2 == 0)
+                    //从有到无显然要经历index1步删除操作
+                    dp[index1][index2] = index1;
+                //三种操作方式
+                //1.word1删除末尾字符 dp[index1-1][index2]+1
+                //2.word1在末尾插入字符 dp[index1][index2-1]+1
+                //3.word1替换末尾字符(如果末尾字符本来就相等,就不用换了) dp[index1-1][index2-1]+(word1[index1-1]!=word2[index2-1]
+                else dp[index1][index2] = min(min(dp[index1-1][index2]+1,dp[index1][index2-1]+1),dp[index1-1][index2-1]+(word1[index1-1]!=word2[index2-1]));
+                //cout << dp[index1][index2] << " ";
+            }
+            //cout << endl;
+        }
+        return dp[len1][len2];
+    }
+};