[Function add]

1. Add leetcode solution.

Author: Seanforfun

leetcode/18. 4Sum.md

@@ -1,6 +1,13 @@
 ## 18. 4Sum
+### Question:
+Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.
+
+Note:
+
+* The solution set must not contain duplicate quadruplets.
+
 ### Thinking:
-* Method: 
+* Method:
 It is very similar to 3sum, but we use one more index to iterate.
 It is very important to remove duplicate elements.
 ```Java
@@ -21,7 +28,7 @@ It is very important to remove duplicate elements.
 	                        while(left < right && nums[left-1] == nums[left])   left++;
 	                        while(left < right && nums[right+1] == nums[right]) right--;
 	                    }else if(temp < target){
-	                        left++;   
+	                        left++;
 	                    }else{
 	                        right--;
 	                    }
@@ -33,4 +40,37 @@ It is very important to remove duplicate elements.
 	        return result;
 	    }
 	}
+```
+
+### 二刷
+1. 和一刷的时候类似，外围二次遍历，内部使用双指针。
+
+```Java
+class Solution {
+    public List<List<Integer>> fourSum(int[] nums, int target) {
+        List<List<Integer>> result = new ArrayList<>();
+        if(nums == null || nums.length < 4) return result;
+        int low = 0, high = 0, sum = 0;
+        Arrays.sort(nums);
+        for(int i = 0; i < nums.length - 3; i++){
+            if(i > 0 && nums[i] == nums[i - 1]) continue;
+            for(int j = i + 1; j < nums.length - 2; j++){
+                if(j - i > 1 && nums[j - 1] == nums[j]) continue;
+                low = j + 1; high = nums.length - 1;
+                while(low < high){
+                    sum = nums[i] + nums[j] + nums[low] + nums[high];
+                    if(sum == target){
+                        result.add(Arrays.asList(nums[i], nums[j], nums[low++], nums[high--]));
+                        while(high > low && nums[high] == nums[high + 1]) high--;
+                        while(low < high && nums[low] == nums[low - 1]) low++;
+                    }else if(sum > target){
+                        --high;
+                    }else
+                        ++low;
+                }
+            }
+        }
+        return result;
+    }
+}
 ```

leetcode/19. Remove Nth Node From End of List.md

@@ -1,6 +1,18 @@
 ## 19. Remove Nth Node From End of List
+
+### Question:
+Given a linked list, remove the n-th node from the end of list and return its head.
+
+```
+Example:
+
+Given linked list: 1->2->3->4->5, and n = 2.
+
+After removing the second node from the end, the linked list becomes 1->2->3->5.
+```
+
 ### Thinking:
-* Method: 
+* Method:
 1. Iterate to get the length of the list.
 2. Iterate again to remove
 3. Trick: Use a dummy node to represent the head.
@@ -38,4 +50,33 @@
 	        return result.next;
 	    }
 	}
+```
+
+### 二刷
+1. 快慢指针，先确定两个指针之间的距离。
+2. 同时向后移动指针，定位出倒数第n个结点。
+3. 通过改变next指针删除元素。
+
+```Java
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public ListNode removeNthFromEnd(ListNode head, int n) {
+        ListNode fast = head, slow = head;
+        for(int i = 0; i < n + 1; i++)
+            fast = fast.next;
+        while(fast != null){
+            fast = fast.next;
+            slow = slow.next;
+        }
+        slow.next = slow.next.next;
+        return head;
+    }
+}
 ```