990与46更新

Author: LuYanFCP

README.md

@@ -4,7 +4,15 @@
 
 ## 分类整理
 
-## 位运算
+### dfs
+
++ No.46 [python](python/46.py)
+
+### 并查集
+
++ No.990 [cpp（未优化）](cpp/990.cpp) [cpp优化](cpp/990_1.cpp)
+
+### 位运算
 
 + No.136 [cpp](cpp/136.cpp)
 + No.64 [cpp](cpp/64.cpp) [python](python/64.py)
@@ -32,6 +40,7 @@
 
 ## 动态规划
 
++ No.46 [cpp]()  [python](python/46_dp.py)
 + No.53 [cpp](cpp/53.cpp) [python](python/53.py)
 + No.120 [cpp](cpp/120.cpp) [python](python/120.py)
 + No.837 [cpp](cpp/837.cpp) [python](python/837.py)

cpp/990.cpp

@@ -0,0 +1,51 @@
+#include <vector>
+#include <string>
+#include <unordered_map>
+using namespace std;
+
+class Solution {
+public:
+    /**
+     * 执行用时 :16 ms, 在所有 C++ 提交中击败了25.68%的用户
+     * 内存消耗 :11.2 MB, 在所有 C++ 提交中击败了50.00%的用户
+    */
+    unordered_map<char, char> eq;
+
+    bool equationsPossible(vector<string>& equations) {
+        int n = equations.size();
+        
+        for(const string &s: equations) { 
+            bool flag1 = insert_node(eq, s[0]);
+            bool flag2 = insert_node(eq, s[3]);  // 插入 
+            if (s[1] == '=') join(eq, s[0], s[3]);
+        }
+
+        for(const string &s: equations) { 
+            if (s[1] == '!') {
+                char root1 = find(eq, s[0]);
+                char root2 = find(eq, s[3]);
+                if (root1 == root2) return false;
+            }
+        }
+
+        return true;
+    }
+
+    bool is_exist(unordered_map<char, char>& set, char x) {
+        return set.find(x) != set.end();
+    }
+    bool insert_node(unordered_map<char, char>& set, char x) {
+        if (is_exist(set, x)) return false;
+        set[x] = x; 
+        return true;
+    }
+    char find(unordered_map<char, char>& set, char x) {
+        if (set[x] == x) return x;
+        return find(set, set[x]);
+    }
+    void join(unordered_map<char, char>& set, char x, char y) {
+        char x_root = find(set, x);
+        char y_root = find(set, y);
+        set[y_root] = x_root;
+    }
+};

cpp/990_1.cpp

@@ -0,0 +1,53 @@
+#include <vector>
+#include <string>
+#include <unordered_map>
+using namespace std;
+
+class Solution {
+public:
+    /**
+     * 提示：
+     * 1. 1 <= equations.length <= 500
+     * 2. equations[i].length == 4
+     * 3. equations[i][0] 和 equations[i][3] 是小写字母
+     * 4. equations[i][1] 要么是 '='，要么是 '!'
+     * 5. equations[i][2] 是 '='
+     */
+    /**
+     * 改进，使用数组代替之前用的hashmap， 增加并查集的路径压缩算法、。
+     * 
+     * 改进后：
+     * 速度： 4 ms, 在所有 C++ 提交中击败了98.83%的用户。
+     * 内存： 内存消耗 :10.8 MB, 在所有 C++ 提交中击败了50.00%的用户。  
+    */
+
+    bool equationsPossible(vector<string>& equations) {
+        // 并查集初始化
+        int set[30];
+        for (int i = 0; i < 30; ++i) set[i] = i;
+        for (const string &s : equations) {
+            if (s[1] == '=') {
+                join(set, s[0]-'a', s[3]-'a');
+            }
+        }
+        for (const string &s : equations) {
+            if (s[1] == '!') {
+                if (find(set, s[0]-'a') == find(set, s[3]-'a'))
+                    return false;
+            }
+        }
+        return true;
+    }
+
+    
+    char find(int *set, int x) {
+        if (set[x] == x) return x;
+        int res = find(set, set[x]);  // 路径压缩算法
+        return res;
+    }
+    void join(int *set, int x, int y) {
+        int root_x = find(set, x);
+        int root_y = find(set, y);
+        set[root_y] = root_x;
+    }
+};

python/46.py

@@ -0,0 +1,35 @@
+class Solution:
+    def translateNum(self, num: int) -> int:
+        # dfs
+        # 12258
+        num_str = str(num)
+        cache = {}  # 缓存， 空间换时间 40ms->36ms
+        # count = 0
+        def _dfs(num_str):
+            # nonlocal count
+            if not len(num_str):
+                # count+=1
+                return 1
+
+            num_str1 = num_str[1:]
+            num_str2 = num_str[2:]
+            
+            if not num_str1 in cache:
+                res1 = _dfs(num_str1)
+                cache[num_str1] = res1
+            else:
+                res1 = cache[num_str1]
+            
+            res2 = 0
+            num_2 = int(num_str[:2])
+            if len(num_str) >= 2 and num_2 <26 and num_2 > 9:
+                if num_str2 in cache:
+                    res2 += _dfs(num_str2)
+                    cache[num_2] = res2
+                else:
+                    res2 = cache[num_str2]
+
+            return res1 + res2
+        
+        # print(count)
+        return _dfs(num_str)

python/46_dp.py

@@ -0,0 +1,26 @@
+class Solution:
+    def translateNum(self, num: int) -> int:
+        str_num = str(num)
+        n = len(num)
+        dp = [0] * n + 1 # 前n个数有多少种可能
+        dp[0] = 1
+        dp[1] = 1
+        for i in range(2, n+1):
+            temp = int(str_num[i-2:i])
+            if temp > 9 and temp < 26:
+                dp[i] += dp[i-2]
+            dp[i] += dp[i-1]
+
+        return dp[n]
+
+    def translateNum2(self, num: int) -> int:
+        n = len(num)
+        str_num = str(num)
+        a, b = 1, 1 # 由于只用到后面两个数，所有可以直接迭代, 节约了空间
+        for i in range(2, n+1):
+            temp = int(str_num[i-2:i])
+            if temp > 9 and temp < 26:
+                a, b = a+b, a
+            else:
+                a, b = a, a
+        return a