rename all the solution and add 242

Author: Li Li

Algorithms Basics/237. Delete Node in a Linked List.cs renamed to Algorithms Basics/237. Delete Node in a Linked List

@@ -0,0 +1,26 @@
+public class Solution {
+    public bool IsAnagram(string s, string t) {
+        if (s == null) return t == null;
+        if (t == null) return false;
+        int m = s.Length, n = t.Length;
+        if (m != n) return false;
+        var dict = new Dictionary<char, int>();
+        for (var i = 0; i < m; i++) {
+            if (!dict.ContainsKey(s[i])) {
+                dict[s[i]] = 1;
+            } else {
+                dict[s[i]]++;
+            }
+        }
+        for (var i = 0; i < n; i++) {
+            if (!dict.ContainsKey(t[i])) {
+                return false;
+            }
+            dict[t[i]]--;
+            if (dict[t[i]] < 0) {
+                return false;
+            }
+        }
+        return true;
+    }
+}

Algorithms Basics/Chapter 1. Array_String/167. Two Sum II - Input array is sorted/167. Two Sum II - Input array is sorted.cs renamed to Algorithms Basics/Chapter 1. Array_String/167. Two Sum II - Input array is sorted/167. Two Sum II - Input array is sorted

@@ -4,7 +4,7 @@ namespace LeetcodeCShaprDotNetCore {
     class Program {
         static void Main(string[] args) {
             var s = new Solution();
-            var r = s.TwoSum(new int[] { 2, 7, 11, 15 }, 9);
+            var r = s.TwoSum(new int[] { 11, 7, 2, 15 }, 9);
             System.Console.WriteLine(r[0] + " " + r[1]);
         }
     }

Algorithms Basics/Chapter 1. Array_String/167. Two Sum II - Input array is sorted/similar problem/1. Two Sum.cs renamed to Algorithms Basics/Chapter 1. Array_String/167. Two Sum II - Input array is sorted/similar problem/1. Two Sum

@@ -4,24 +4,29 @@
 
 namespace LeetcodeCShaprDotNetCore {
     public class Solution {
-        public int[] TwoSum(int[] nums, int target) {
-            int[] res = null;
-            if (nums == null || nums.Length == 0) {
-                return res;
+        public bool IsAnagram(string s, string t) {
+            if (s == null) return t == null;
+            if (t == null) return false;
+            int m = s.Length, n = t.Length;
+            if (m != n) return false;
+            var dict = new Dictionary<char, int>();
+            for (var i = 0; i < m; i++) {
+                if (!dict.ContainsKey(s[i])) {
+                    dict[s[i]] = 1;
+                } else {
+                    dict[s[i]]++;
+                }
             }
-            for (int i = 0, j = nums.Length - 1; i < j;) {
-                var v = nums[i] + nums[j];
-                if (v == target) {
-                    res = new int[] { i + 1, j + 1 };
-                    break;
+            for (var i = 0; i < n; i++) {
+                if (!dict.ContainsKey(t[i])) {
+                    return false;
                 }
-                if (v < target) {
-                    i++;
-                } else {
-                    j--;
+                dict[t[i]]--;
+                if (dict[t[i]] < 0) {
+                    return false;
                 }
             }
-            return res;
+            return true;
         }
     }
 }

Algorithms Basics/Chapter 1. Array_String/167. Two Sum II - Input array is sorted/similar problem/653. Two Sum IV - Input is a BST.cs renamed to Algorithms Basics/Chapter 1. Array_String/167. Two Sum II - Input array is sorted/similar problem/653. Two Sum IV - Input is a BST

@@ -55,4 +55,12 @@ Since the array is already sorted, we can keep two pointers *left* and *right*.
 
 Similar problems:
 *1.* Two Sum (use dict for each item just use almost O(1) check time, so totally O(n) time);
-*653.* Two Sum IV - Input is a BST (3 solution to revisit) 
+*653.* Two Sum IV - Input is a BST (3 solution to revisit) 
+
+#### 242. Valid Anagram
+
+This problem can be solved by using Dictionary.
+
+**Algorithm**
+First use dict record the counts for each char in string1, then check for string2, for each char c, if !dict.ContainsKey(c), return fasle; else dict[c]--, if dict[c] < 0, return false.
+