Merge pull request #93 from HuizeHuang/master

第一周作业-#48

Author: GeekUniversity

Week_01/id_48/LeetCode_236_48.java

@@ -0,0 +1,65 @@
+//Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
+//
+// According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”
+//
+// Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]
+//
+//
+//
+// Example 1:
+//
+//
+//Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
+//Output: 3
+//Explanation: The LCA of nodes 5 and 1 is 3.
+//
+//
+// Example 2:
+//
+//
+//Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
+//Output: 5
+//Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.
+//
+//
+//
+//
+// Note:
+//
+//
+// All of the nodes' values will be unique.
+// p and q are different and both values will exist in the binary tree.
+//
+//
+
+import javax.swing.tree.TreeNode;
+
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
+        /**
+         * 摘自解答：
+         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存在于左或右子树中, 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 == null || root == p || root == q) return root;
+        TreeNode left = lowestCommonAncestor(root.left, p, q);
+        TreeNode right = lowestCommonAncestor(root.right, p, q);
+        if (left != null && right != null) return root;
+        else if (left == null && right == null) return null;
+        else return left == null ? right : left;
+
+    }
+}

Week_01/id_48/LeetCode_26_48.java

@@ -0,0 +1,56 @@
+//Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length.
+//
+// Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
+//
+// Example 1:
+//
+//
+//Given nums = [1,1,2],
+//
+//Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.
+//
+//It doesn't matter what you leave beyond the returned length.
+//
+// Example 2:
+//
+//
+//Given nums = [0,0,1,1,1,2,2,3,3,4],
+//
+//Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.
+//
+//It doesn't matter what values are set beyond the returned length.
+//
+//
+// Clarification:
+//
+// Confused why the returned value is an integer but your answer is an array?
+//
+// Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.
+//
+// Internally you can think of this:
+//
+//
+//// nums is passed in by reference. (i.e., without making a copy)
+//int len = removeDuplicates(nums);
+//
+//// any modification to nums in your function would be known by the caller.
+//// using the length returned by your function, it prints the first len elements.
+//for (int i = 0; i < len; i++) {
+//    print(nums[i]);
+//}
+//
+
+class Solution {
+    public int removeDuplicates(int[] nums) {
+        // two pointers
+        if (nums == null || nums.length == 0) return 0;
+        // i is the last non-duplicated digit, j is fast pointer
+        int i = 0;
+        for (int j = 1; j < nums.length; j++){
+            if (nums[j] != nums[i]) {
+                nums[++i] = nums[j];
+            }
+        }
+        return i + 1;
+    }
+}