20190303

Author: mJackie

Diff for: code/lc138.java

@@ -31,7 +31,7 @@ public RandomListNode copyRandomList(RandomListNode head) {
         node = head;
         RandomListNode res = head.next;
         RandomListNode temp = head.next;
-        while(temp.next!=null){
+        while(temp!=null&&temp.next!=null){
             node.next = temp.next;
             temp.next = node.next.next;
             node = node.next;

Diff for: code/lc242.java

@@ -4,8 +4,8 @@
  * 题意：字符串t是否为s打乱后的重排列
  * 难度：Easy
  * 分类：Hash Table, Sort
- * 思路：
- * Tips：
+ * 思路：hash table 计算即可，<0了直接返回false
+ * Tips：和重排列那题联系一下 lc46
  */
 public class lc242 {
     public boolean isAnagram(String s, String t) {

Diff for: code/lc328.java

@@ -0,0 +1,32 @@
+package code;
+/*
+ * 328. Odd Even Linked List
+ * 题意：奇数的node都在偶数node的后面，第一个节点index为0
+ * 难度：Medium
+ * 分类：Linked List
+ * 思路：while的条件很不好想，记住：这种跳两次的迭代，都判断后边那个节点的 next!=null 为终止条件
+ * Tips：lc318
+ */
+public class lc328 {
+    public class ListNode {
+        int val;
+        ListNode next;
+        ListNode(int x) {
+            val = x;
+        }
+    }
+    public ListNode oddEvenList(ListNode head) {
+        if(head==null||head.next==null) return head;
+        ListNode even = head;
+        ListNode odd_head = head.next;
+        ListNode odd = head.next;
+        while( odd!=null && odd.next!=null ){   //这个条件很难想通
+            even.next = even.next.next;
+            odd.next = odd.next.next;
+            even = even.next;
+            odd = odd.next;
+        }
+        even.next = odd_head;
+        return head;
+    }
+}

Diff for: code/lc334.java

@@ -0,0 +1,21 @@
+package code;
+/*
+ * 334. Increasing Triplet Subsequence
+ * 题意：数组中是否存在递增的3个数
+ * 难度：Medium
+ * 分类：Array
+ * 思路：思路很清奇，自己没想到，时间空间复杂度都是O(1)。其实和lc300 O(nlgn) 解法思路是一样的，只是固定了dp数组长度为两个数。
+ * Tips：lc300最长递增子序列
+ */
+public class lc334 {
+    public boolean increasingTriplet(int[] nums) {
+        int small = Integer.MAX_VALUE;
+        int big = Integer.MAX_VALUE;
+        for (int i = 0; i < nums.length ; i++) {
+            if(nums[i]<small) small = nums[i];
+            else if(nums[i]<big) big = nums[i];
+            else return true;
+        }
+        return false;
+    }
+}

Diff for: code/lc341.java

@@ -0,0 +1,51 @@
+package code;
+
+import java.util.Iterator;
+import java.util.List;
+import java.util.Stack;
+
+public class lc341 {
+     public interface NestedInteger {
+         // @return true if this NestedInteger holds a single integer, rather than a nested list.
+         public boolean isInteger();
+         // @return the single integer that this NestedInteger holds, if it holds a single integer
+         // Return null if this NestedInteger holds a nested list
+         public Integer getInteger();
+         // @return the nested list that this NestedInteger holds, if it holds a nested list
+         // Return null if this NestedInteger holds a single integer
+          public List<NestedInteger> getList();
+    }
+    /**
+     * Your NestedIterator object will be instantiated and called as such:
+     * NestedIterator i = new NestedIterator(nestedList);
+     * while (i.hasNext()) v[f()] = i.next();
+     */
+    public class NestedIterator implements Iterator<Integer> {
+        Stack<NestedInteger> stack = new Stack<>();
+        public NestedIterator(List<NestedInteger> nestedList) {
+            for(int i = nestedList.size() - 1; i >= 0; i--) {
+                stack.push(nestedList.get(i));
+            }
+        }
+
+        @Override
+        public Integer next() {
+            return stack.pop().getInteger();
+        }
+
+        @Override
+        public boolean hasNext() {
+            while(!stack.isEmpty()) {
+                NestedInteger curr = stack.peek();
+                if(curr.isInteger()) {
+                    return true;
+                }
+                stack.pop();
+                for(int i = curr.getList().size() - 1; i >= 0; i--) {
+                    stack.push(curr.getList().get(i));
+                }
+            }
+            return false;
+        }
+    }
+}

Diff for: code/lc347.java

@@ -22,7 +22,7 @@ public static List<Integer> topKFrequent(int[] nums, int k) {
         for (int i = 0; i < nums.length ; i++) {    //放入hashmap计数
             hm.put(nums[i], hm.getOrDefault(nums[i], 0)+1);
         }
-        for( int i : hm.keySet() ){ //key,value反转，放入treemap
+        for( int i : hm.keySet() ){ //key,value反转，放入treemap  TreeMap中默认是按照升序进行排序的
             int freq = hm.get(i);
             if(tm.containsKey(freq))
                 tm.get(freq).add(i);

Diff for: code/lc378.java

@@ -0,0 +1,14 @@
+package code;
+/*
+ * 378. Kth Smallest Element in a Sorted Matrix
+ * 题意：在矩阵中搜索第k大的数，横轴和纵轴都是有序的
+ * 难度：Medium
+ * 分类：Binary Search, Heap
+ * 思路：
+ * Tips：lc23方法很像
+ */
+public class lc378 {
+    public int kthSmallest(int[][] matrix, int k) {
+        
+    }
+}

Diff for: readme.md

@@ -127,9 +127,12 @@ Language: Java
 | 227 [Java](./code/lc227.java)
 | 230 [Java](./code/lc230.java)
 | 234 [Java](./code/lc234.java)
+| 237 [Java](./code/lc237.java)
 | 238 [Java](./code/lc238.java)
 | 239 [Java](./code/lc239.java)
 | 240 [Java](./code/lc240.java)
+| 242 [Java](./code/lc242.java)
+| 268 [Java](./code/lc268.java)
 | 279 [Java](./code/lc279.java)
 | 283 [Java](./code/lc283.java)
 | 287 [Java](./code/lc287.java)
@@ -141,6 +144,9 @@ Language: Java
 | 312 [Java](./code/lc312.java)
 | 315 [Java](./code/lc315.java)
 | 322 [Java](./code/lc322.java)
+| 324 [Java](./code/lc324.java)
+| 326 [Java](./code/lc326.java)
+| 329 [Java](./code/lc329.java)
 | 337 [Java](./code/lc337.java)
 | 338 [Java](./code/lc338.java)
 | 347 [Java](./code/lc347.java)