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01.计算机科学/算法/leetcode/1184.公交站间的距离.cs

@@ -0,0 +1,90 @@
+/*
+ * @lc app=leetcode.cn id=1184 lang=csharp
+ *
+ * [1184] 公交站间的距离
+ *
+ * https://leetcode-cn.com/problems/distance-between-bus-stops/description/
+ *
+ * algorithms
+ * Easy (57.84%)
+ * Likes:    39
+ * Dislikes: 0
+ * Total Accepted:    10.5K
+ * Total Submissions: 18.1K
+ * Testcase Example:  '[1,2,3,4]\n0\n1'
+ *
+ * 环形公交路线上有 n 个站，按次序从 0 到 n - 1 进行编号。我们已知每一对相邻公交站之间的距离，distance[i] 表示编号为 i
+ * 的车站和编号为 (i + 1) % n 的车站之间的距离。
+ * 
+ * 环线上的公交车都可以按顺时针和逆时针的方向行驶。
+ * 
+ * 返回乘客从出发点 start 到目的地 destination 之间的最短距离。
+ * 
+ * 
+ * 
+ * 示例 1：
+ * 
+ * 
+ * 
+ * 输入：distance = [1,2,3,4], start = 0, destination = 1
+ * 输出：1
+ * 解释：公交站 0 和 1 之间的距离是 1 或 9，最小值是 1。
+ * 
+ * 
+ * 
+ * 示例 2：
+ * 
+ * 
+ * 
+ * 输入：distance = [1,2,3,4], start = 0, destination = 2
+ * 输出：3
+ * 解释：公交站 0 和 2 之间的距离是 3 或 7，最小值是 3。
+ * 
+ * 
+ * 
+ * 
+ * 示例 3：
+ * 
+ * 
+ * 
+ * 输入：distance = [1,2,3,4], start = 0, destination = 3
+ * 输出：4
+ * 解释：公交站 0 和 3 之间的距离是 6 或 4，最小值是 4。
+ * 
+ * 
+ * 
+ * 
+ * 提示：
+ * 
+ * 
+ * 1 <= n <= 10^4
+ * distance.length == n
+ * 0 <= start, destination < n
+ * 0 <= distance[i] <= 10^4
+ * 
+ * 
+ */
+
+// @lc code=start
+public class Solution {
+    public int DistanceBetweenBusStops (int[] distance, int start, int destination) {
+        if (start == destination) return 0;
+        int min = Math.Min (start, destination);
+        int max = Math.Max (start, destination);
+        int dis1 = 0;
+        int dis2 = 0;
+        for (int i = min; i < max; i++) {
+            dis1 += distance[i];
+        }
+
+        int j = max;
+        while (max != min) {
+            if (j < distance.Length - 1) {
+                j++;
+            } else if (j == distance.Length - 1) {
+                j = 0;
+            }
+        }
+    }
+}
+// @lc code=end

01.计算机科学/算法/leetcode/133.克隆图.cs

@@ -111,9 +111,22 @@ public Node(int _val, List<Node> _neighbors) {
 */
 
 public class Solution {
-    public Node CloneGraph(Node node) {
-
+    //TODO:
+    public Node CloneGraph (Node node) {
+        if (node == null) return node;
+        Dictionary<Node, Node> dic = new Dictionary<Node, Node> ();
+        return dfs (node, dic);
     }
-}
-// @lc code=end
 
+    private Node dfs (Node node, Dictionary<Node, Node> dic) {
+        if (node == null) return node;
+        if (dic.ContainsKey (node)) return dic[node];
+        Node copy = new Node (node.val);
+        dic.Add (node, copy);
+        foreach (var item in node.neighbors) {
+            copy.neighbors.Add (dfs (item, dic));
+        }
+        return copy;
+    }
+}
+// @lc code=end

01.计算机科学/算法/leetcode/138.复制带随机指针的链表.cs

@@ -0,0 +1,109 @@
+/*
+ * @lc app=leetcode.cn id=138 lang=csharp
+ *
+ * [138] 复制带随机指针的链表
+ *
+ * https://leetcode-cn.com/problems/copy-list-with-random-pointer/description/
+ *
+ * algorithms
+ * Medium (57.71%)
+ * Likes:    414
+ * Dislikes: 0
+ * Total Accepted:    49.8K
+ * Total Submissions: 86.2K
+ * Testcase Example:  '[[7,null],[13,0],[11,4],[10,2],[1,0]]'
+ *
+ * 给定一个链表，每个节点包含一个额外增加的随机指针，该指针可以指向链表中的任何节点或空节点。
+ * 
+ * 要求返回这个链表的 深拷贝。 
+ * 
+ * 我们用一个由 n 个节点组成的链表来表示输入/输出中的链表。每个节点用一个 [val, random_index] 表示：
+ * 
+ * 
+ * val：一个表示 Node.val 的整数。
+ * random_index：随机指针指向的节点索引（范围从 0 到 n-1）；如果不指向任何节点，则为  null 。
+ * 
+ * 
+ * 
+ * 
+ * 示例 1：
+ * 
+ * 
+ * 
+ * 输入：head = [[7,null],[13,0],[11,4],[10,2],[1,0]]
+ * 输出：[[7,null],[13,0],[11,4],[10,2],[1,0]]
+ * 
+ * 
+ * 示例 2：
+ * 
+ * 
+ * 
+ * 输入：head = [[1,1],[2,1]]
+ * 输出：[[1,1],[2,1]]
+ * 
+ * 
+ * 示例 3：
+ * 
+ * 
+ * 
+ * 输入：head = [[3,null],[3,0],[3,null]]
+ * 输出：[[3,null],[3,0],[3,null]]
+ * 
+ * 
+ * 示例 4：
+ * 
+ * 输入：head = []
+ * 输出：[]
+ * 解释：给定的链表为空（空指针），因此返回 null。
+ * 
+ * 
+ * 
+ * 
+ * 提示：
+ * 
+ * 
+ * -10000 <= Node.val <= 10000
+ * Node.random 为空（null）或指向链表中的节点。
+ * 节点数目不超过 1000 。
+ * 
+ * 
+ */
+
+// @lc code=start
+/*
+// Definition for a Node.
+public class Node {
+    public int val;
+    public Node next;
+    public Node random;
+    
+    public Node(int _val) {
+        val = _val;
+        next = null;
+        random = null;
+    }
+}
+*/
+
+public class Solution {
+    public Node CopyRandomList (Node head) {
+        if (node == null) return node;
+        Dictionary<Node, List<Node>> dic = new Dictionary<Node, List<Node>> ();
+        return dfs (node, dic);
+    }
+
+    private Node dfs (Node node, Dictionary<Node, List<Node>> dic) {
+        if (dic.ContainsKey (node)) return dic[node];
+        Node copy = new Node (node.val);
+        dic.Add (node, new List<Node> ());
+        if (node.next != null) {
+            dic[node].next = dfs (node.next, dic);
+        }
+
+        if (node.random != null) {
+            dic[node].random = dfs (node.random, dic);
+        }
+        return copy;
+    }
+}
+// @lc code=end

01.计算机科学/算法/leetcode/49.字母异位词分组.cs

@@ -0,0 +1,70 @@
+/*
+ * @lc app=leetcode.cn id=49 lang=csharp
+ *
+ * [49] 字母异位词分组
+ *
+ * https://leetcode-cn.com/problems/group-anagrams/description/
+ *
+ * algorithms
+ * Medium (63.90%)
+ * Likes:    496
+ * Dislikes: 0
+ * Total Accepted:    114.5K
+ * Total Submissions: 179.2K
+ * Testcase Example:  '["eat","tea","tan","ate","nat","bat"]'
+ *
+ * 给定一个字符串数组，将字母异位词组合在一起。字母异位词指字母相同，但排列不同的字符串。
+ * 
+ * 示例:
+ * 
+ * 输入: ["eat", "tea", "tan", "ate", "nat", "bat"]
+ * 输出:
+ * [
+ * ⁠ ["ate","eat","tea"],
+ * ⁠ ["nat","tan"],
+ * ⁠ ["bat"]
+ * ]
+ * 
+ * 说明：
+ * 
+ * 
+ * 所有输入均为小写字母。
+ * 不考虑答案输出的顺序。
+ * 
+ * 
+ */
+
+// @lc code=start
+public class Solution {
+    public IList<IList<string>> GroupAnagrams (string[] strs) {
+        Dictionary<string, IList<string>> dic = new Dictionary<string, IList<string>> ();
+        foreach (var str in strs) {
+            int[] cs = new int[26];
+            foreach (var c in str) {
+                cs[c - 'a']++;
+            }
+            StringBuilder sb = new StringBuilder ();
+            for (int i = 0; i < 26; i++) {
+                if (cs[i] > 0) {
+                    sb.Append ((char) ('a' + i));
+                    sb.Append (cs[i]);
+                }
+            }
+            string s = sb.ToString ();
+            if (dic.ContainsKey (s)) {
+                dic[s].Add (str);
+            } else {
+                List<string> list = new List<string> ();
+                list.Add (str);
+                dic.Add (s, list.ToList ());
+            }
+        }
+        IList<IList<string>> res = new List<IList<string>> ();
+        foreach (var item in dic) {
+            var list = item.Value;
+            res.Add (list);
+        }
+        return res;
+    }
+}
+// @lc code=end

01.计算机科学/算法/leetcode/56.合并区间.cs

@@ -44,42 +44,64 @@
 
 // @lc code=start
 public class Solution {
-    //分治算法
+    //TODO:
     public int[][] Merge (int[][] intervals) {
-        List<int[]> result = new List<int[]> ();
-        if(intervals.Length == 0)return result.ToArray();
-        for(int i = 0;i<intervals.Length - 1;i++){
-            var a = intervals[i];
-            bool has = false;
-            for(int j = i+1;j<intervals.Length;j++){
-                var b = intervals[j];
-                if(CanMerge(a,b)){
-                    a = MergeFunc1 (a, b);
-                    has = true;
+        if (intervals.Length == 0) {
+            return intervals;
+        }
+
+        intervals = intervals.OrderBy (p => p[0]).ToArray ();
+        List<int[]> list = new List<int[]> ();
+        for (int i = 0; i < intervals.Length - 1; i++) {
+            if (intervals[i][1] >= intervals[i + 1][0]) {
+                intervals[i + 1][0] = intervals[i][0];
+                if (intervals[i][1] >= intervals[i + 1][1]) {
+                    intervals[i + 1][1] = intervals[i][1];
                 }
-            }
-            if(has == false){
-                result.Add (a);
+            } else {
+                list.Add (intervals[i]);
             }
         }
-        return result.ToArray();
+        list.Add (intervals[intervals.Length - 1]);
+        int[][] result = list.ToArray ();
+        return result;
     }
 
-    private bool CanMerge (int[] a, int[] b) {
-        if (a == null || b == null) {
-            return true;
-        } else if (a[1] < b[0]) {
-            return false;
-        } else if (b[1] < a[0]) {
-            return false;
-        }
-        return true;
-    }
+    // private void Func (int[][] intervals) {
+    //     List<int[]> result = new List<int[]> ();
+    //     if (intervals.Length == 0) return result.ToArray ();
+    //     for (int i = 0; i < intervals.Length - 1; i++) {
+    //         var a = intervals[i];
+    //         bool has = false;
+    //         for (int j = i + 1; j < intervals.Length; j++) {
+    //             var b = intervals[j];
+    //             if (CanMerge (a, b)) {
+    //                 a = MergeFunc1 (a, b);
+    //                 has = true;
+    //             }
+    //         }
+    //         if (has == false) {
+    //             result.Add (a);
+    //         }
+    //     }
+    //     return result.ToArray ();
+    // }
 
-    private int[] MergeFunc1 (int[] a, int[] b) {
-        int min = Math.Min (a[0], b[0]);
-        int max = Math.Max (a[1], b[1]);
-        return new int[2] { min, max };
-    }
+    // private bool CanMerge (int[] a, int[] b) {
+    //     if (a == null || b == null) {
+    //         return true;
+    //     } else if (a[1] < b[0]) {
+    //         return false;
+    //     } else if (b[1] < a[0]) {
+    //         return false;
+    //     }
+    //     return true;
+    // }
+
+    // private int[] MergeFunc1 (int[] a, int[] b) {
+    //     int min = Math.Min (a[0], b[0]);
+    //     int max = Math.Max (a[1], b[1]);
+    //     return new int[2] { min, max };
+    // }
 }
 // @lc code=end