From 2d9f4774f0359f5233293e3d7e4e675a514ecf2e Mon Sep 17 00:00:00 2001
From: liu13 <1099976891@qq.com>
Date: Wed, 20 Mar 2019 10:14:26 +0800
Subject: [PATCH] 20190320

---
 code/lc112.java | 30 ++++++++++++++++++++++++++++++
 code/lc113.java |  2 +-
 code/lc124.java |  2 +-
 code/lc129.java | 28 ++++++++++++++++++++++++++++
 code/lc151.java | 19 +++++++++++++++++++
 code/lc236.java |  6 +++---
 code/lc239.java |  6 +++---
 code/lc268.java |  4 ++--
 code/lc279.java |  2 +-
 code/lc287.java |  3 ++-
 code/lc303.java |  1 +
 code/lc337.java |  3 ++-
 code/lc378.java |  1 +
 code/lc395.java |  2 +-
 code/lc416.java |  3 ++-
 code/lc437.java | 26 ++++----------------------
 code/lc448.java |  2 +-
 code/lc494.java |  5 ++++-
 code/lc538.java | 13 +++++--------
 code/lc543.java |  2 +-
 code/lc560.java |  1 +
 code/lc617.java |  8 +++-----
 22 files changed, 116 insertions(+), 53 deletions(-)
 create mode 100644 code/lc112.java
 create mode 100644 code/lc129.java
 create mode 100644 code/lc151.java

diff --git a/code/lc112.java b/code/lc112.java
new file mode 100644
index 0000000..c3bdd82
--- /dev/null
+++ b/code/lc112.java
@@ -0,0 +1,30 @@
+package code;
+/*
+ * 112. Path Sum
+ * 题意：是否存在跟到叶子节点的和为sum
+ * 难度：Easy
+ * 分类：
+ * 思路：
+ * Tips：lc112, lc113, lc437, lc129, lc124, lc337
+ * 总结一下 lc112 找跟到叶子和为sum的路径是否存在
+ *         lc113 把lc112的路径找出来
+ *         lc437 是任意一条向下走的路径
+ *         lc129 计算所有的和
+ *         lc124 任意一条路径
+ *         lc337 树的dp
+ */
+public class lc112 {
+    public class TreeNode {
+        int val;
+        TreeNode left;
+        TreeNode right;
+        TreeNode(int x) {
+            val = x;
+        }
+    }
+    public boolean hasPathSum(TreeNode root, int sum) {
+        if(root==null) return false;
+        if(root.val==sum&&root.left==null&&root.right==null) return true;
+        return hasPathSum(root.left, sum-root.val)||hasPathSum(root.right, sum-root.val);
+    }
+}
diff --git a/code/lc113.java b/code/lc113.java
index 6d555e2..b8515e0 100644
--- a/code/lc113.java
+++ b/code/lc113.java
@@ -8,7 +8,7 @@ import java.util.List;
  * 难度：Medium
  * 分类：Tree, Depth-first Search
  * 思路：回溯，注意因为节点上可能正值，可能负值，所以不能剪枝
- * Tips：lc124
+ * Tips：lc112, lc113, lc437, lc129, lc124, lc337
  */
 public class lc113 {
     public class TreeNode {
diff --git a/code/lc124.java b/code/lc124.java
index 3546ba7..2548775 100644
--- a/code/lc124.java
+++ b/code/lc124.java
@@ -6,7 +6,7 @@ package code;
  * 分类：Tree, Depth-first Search
  * 思路：因为二叉树只有两个节点，一条路径可以想象成倒V字，从低层的某个节点一路向上，到达一个顶点，再一路向下，理解了这一点，整道题就好解了。
  * Tips：用了一个全局变量存储最后结果，因为函数返回的是直线路径上的最优解，而不是V字路径最优解
- *      lc133
+ *      lc112, lc113, lc437, lc129, lc124, lc337, lc543
  */
 public class lc124 {
     public class TreeNode {
diff --git a/code/lc129.java b/code/lc129.java
new file mode 100644
index 0000000..736ddb4
--- /dev/null
+++ b/code/lc129.java
@@ -0,0 +1,28 @@
+package code;
+/*
+ * 129. Sum Root to Leaf Numbers
+ * 题意：一条路径上的数字组成一个数字，求所有从跟到叶子的路径和
+ * 难度：Medium
+ * 分类：Tree, Depth-first Search
+ * 思路：dfs
+ * Tips：lc112, lc113, lc437, lc129, lc124, lc337
+ */
+public class lc129 {
+    public class TreeNode {
+        int val;
+        TreeNode left;
+        TreeNode right;
+        TreeNode(int x) {
+            val = x;
+        }
+    }
+    public int sumNumbers(TreeNode root) {
+        if(root==null) return 0;
+        return helper(root, 0);
+    }
+    public int helper(TreeNode root, int sum){
+        if(root==null) return 0;    //一条路径，另一边返回0
+        if(root.left==null&&root.right==null) return sum*10+root.val;   //这点注意，是叶子节点直接返回了
+        return helper(root.left, sum*10+root.val) + helper(root.right, sum*10+root.val);
+    }
+}
diff --git a/code/lc151.java b/code/lc151.java
new file mode 100644
index 0000000..9b7f9f0
--- /dev/null
+++ b/code/lc151.java
@@ -0,0 +1,19 @@
+package code;
+
+import java.util.Arrays;
+import java.util.Collections;
+/*
+ * 151. Reverse Words in a String
+ * 题意：反转字符串中的单词
+ * 难度：Medium
+ * 分类：String
+ * 思路：难点在中间的空格可能是多个空格，注意如何解决
+ * Tips：
+ */
+public class lc151 {
+    public String reverseWords(String s) {
+        String[] words = s.trim().split(" +"); //+号匹配多个
+        Collections.reverse(Arrays.asList(words));
+        return String.join(" ", words);
+    }
+}
diff --git a/code/lc236.java b/code/lc236.java
index 24539b1..fb0b274 100644
--- a/code/lc236.java
+++ b/code/lc236.java
@@ -35,7 +35,7 @@ public class lc236 {
         parent.put(root, null);
         stack.push(root);
 
-        while (!parent.containsKey(p) || !parent.containsKey(q)) {
+        while (!parent.containsKey(p) || !parent.containsKey(q)) {  //遍历了一遍节点，把节点的父节点信息记录了一下
             TreeNode node = stack.pop();
             if (node.left != null) {
                 parent.put(node.left, node);
@@ -47,11 +47,11 @@ public class lc236 {
             }
         }
         Set<TreeNode> ancestors = new HashSet<>();
-        while (p != null) {
+        while (p != null) {     //p的路径节点添加到hashset中
             ancestors.add(p);
             p = parent.get(p);
         }
-        while (!ancestors.contains(q))
+        while (!ancestors.contains(q)) //第一个hashset中遇到的节点，就是最近公共祖先
             q = parent.get(q);
         return q;
     }
diff --git a/code/lc239.java b/code/lc239.java
index 1b33167..3044d85 100644
--- a/code/lc239.java
+++ b/code/lc239.java
@@ -1,10 +1,10 @@
 package code;
 /*
  * 239. Sliding Window Maximum
- * 题意：滑动窗口最大值
+ * 题意：滑动窗口中最大值
  * 难度：Hard
  * 分类：Heap
- * 思路：用双向队列，保证队列里是递增的。单调队列，好好学习一下。
+ * 思路：用双向队列，保证队列里是递减的。单调队列，好好学习一下。
  * Tips：与lc84做比较，84是递增栈
  */
 import java.util.ArrayDeque;
@@ -25,7 +25,7 @@ public class lc239 {
             return new int[]{};
         int[] res = new int[nums.length-k+1];
         int cur = 0;
-        Deque<Integer> dq = new ArrayDeque();
+        Deque<Integer> dq = new ArrayDeque();   //队列里是递减的
         for (int i = 0; i < nums.length ; i++) {
             if( !dq.isEmpty() && dq.peekFirst()<=i-k)
                 dq.removeFirst();
diff --git a/code/lc268.java b/code/lc268.java
index abd2552..30e4f4f 100644
--- a/code/lc268.java
+++ b/code/lc268.java
@@ -7,7 +7,7 @@ package code;
  * 思路：两种巧妙的方法，时间空间都是O(1)
  *      异或
  *      求和以后，减去所有
- * Tips：
+ * Tips：lc268 lc448 lc287
  */
 public class lc268 {
     public int missingNumber(int[] nums) {
@@ -16,7 +16,7 @@ public class lc268 {
         return res;
     }
     public int missingNumber2(int[] nums) {
-        int res = nums.length;
+        int res = nums.length;  //异或上长度
         for(int i=0; i<nums.length; i++) res^=i^nums[i];
         return res;
     }
diff --git a/code/lc279.java b/code/lc279.java
index 62a4f59..32ac5a6 100644
--- a/code/lc279.java
+++ b/code/lc279.java
@@ -17,7 +17,7 @@ public class lc279 {
         int[] dp = new int[n];
         Arrays.fill(dp,Integer.MAX_VALUE);
         for (int i = 1; i <= n ; i++) { //两个for循环
-            for (int j=1; j<=i ; j++) {
+            for (int j=1; j*j<=i ; j++) {
                 if(j*j==i)
                     dp[i-1] = 1;
                 if(j*j<i){
diff --git a/code/lc287.java b/code/lc287.java
index 9a4bbeb..ab8ae7c 100644
--- a/code/lc287.java
+++ b/code/lc287.java
@@ -3,11 +3,12 @@ package code;
  * 287. Find the Duplicate Number
  * 题意：n+1个数属于[1~n]，找出重复的那个数
  * 难度：Medium
- * 分类：Array, Two Pointers, Binary Search
+ * 分类：Array, Two Pointers, Binary Searn+1个数属于[1~n]，找出重复的那个数ch
  * 思路：如果nums[i]不在对应位置，则和对应位置交换。如果对应位置上也为该数，说明这个数就是重复的数字。这个方法改变了数组。是错误的。
  *      另一种方法，把问题转换成有环链表，找环的起始节点。O(n) O(1) lc142
  *      二分查找，每次看一边数字的个数, O(nlog(n)) O(1)
  * Tips：剑指offer原题
+ *       lc268 lc448 lc287
  */
 public class lc287 {
     public int findDuplicate(int[] nums) {  //该方法修改了数组，是错误的，没看清题意
diff --git a/code/lc303.java b/code/lc303.java
index 2ae3820..cccc16f 100644
--- a/code/lc303.java
+++ b/code/lc303.java
@@ -6,6 +6,7 @@ package code;
  * 分类：Dynamic Programming
  * 思路：
  * Tips：Bingo!
+ *      lc303, lc437, lc560
  */
 public class lc303 {
     class NumArray {
diff --git a/code/lc337.java b/code/lc337.java
index 3031019..9f78f4d 100644
--- a/code/lc337.java
+++ b/code/lc337.java
@@ -11,6 +11,7 @@ import java.util.HashMap;
  * Tips：https://leetcode.com/problems/house-robber-iii/discuss/79330/Step-by-step-tackling-of-the-problem
  *       解答给了三种方法，递归，记忆递归，dp
  *       多理解一下树的dp，核心是返回数组而不是一个数字
+ *       lc112, lc113, lc437, lc129, lc124, lc337
  */
 public class lc337 {
     public class TreeNode {
@@ -26,7 +27,7 @@ public class lc337 {
     public int helper(TreeNode root, HashMap<TreeNode, Integer> mem){
         if(root==null)
             return 0;
-        if(mem.containsKey(root))   //用mem去记忆一下资情况的结果，防止重复计算
+        if(mem.containsKey(root))   //用mem去记忆一下子情况的结果，防止重复计算
             return mem.get(root);
         int val =0;
         if(root.left!=null){
diff --git a/code/lc378.java b/code/lc378.java
index e9d398e..07076ef 100644
--- a/code/lc378.java
+++ b/code/lc378.java
@@ -11,6 +11,7 @@ import java.util.PriorityQueue;
  * 思路：两种思路。 1是类似多个有序链表合并的思路，优先队列。
  *                2是二分，二分的是val，看比这个val小的数是不是k
  * Tips：lc23方法很像
+ *       lc240
  */
 public class lc378 {
     class Cell{
diff --git a/code/lc395.java b/code/lc395.java
index 1270e8e..2867448 100644
--- a/code/lc395.java
+++ b/code/lc395.java
@@ -23,7 +23,7 @@ public class lc395 {
 
                 if( cur_uni_char==i && less_than_k_char==i) res = Math.max(res, right-left);
 
-                else if(cur_uni_char>i){    //左边推进
+                else if(cur_uni_char>i){    //左边推进。不在外边加上一个循环的话，就不知道怎么推荐左指针了。
                     while(cur_uni_char!=i){
                         map[s.charAt(left)-'a']--;
                         if(map[s.charAt(left)-'a']==0) cur_uni_char--;
diff --git a/code/lc416.java b/code/lc416.java
index 896aad1..ee1d578 100644
--- a/code/lc416.java
+++ b/code/lc416.java
@@ -9,7 +9,8 @@ import java.util.HashSet;
  * 思路：题意可以转换为用任意个元素组成的和等于数组和/2。可以和 lc1, lc15 3-Sum 对比。
  *      0，1背包问题，递推比较简单，所以空间可以压缩成一维
  *      自己想的思路其实和压缩后的0，1背包类似，但没想到该问题可以抽象为0，1背包
- * Tips：
+ *       dp[i][j] = dp[i-1][j] || dp[i-1][j-nums[i]]
+ * Tips：lc416, lc494
  */
 public class lc416 {
     public static void main(String[] args) {
diff --git a/code/lc437.java b/code/lc437.java
index 1bf0e30..36a8326 100644
--- a/code/lc437.java
+++ b/code/lc437.java
@@ -12,35 +12,19 @@ import java.util.HashMap;
  *       递归的时候用减是否==0的方式，而不是+==sum的方式
  *       和lc560有共同的思想，每个节点只需遍历一遍就可以了
  *       虽然是Easy题，做好也不简单
+ *       lc112, lc113, lc437, lc129, lc124, lc337
+ *       lc303, lc437, lc560
  */
 public class lc437 {
     public static class TreeNode {
         int val;
         TreeNode left;
         TreeNode right;
-
         TreeNode(int x) {
             val = x;
         }
     }
 
-    public static void main(String[] args) {
-        TreeNode node10 = new TreeNode(10);
-        TreeNode node5 = new TreeNode(5);
-        TreeNode node3 = new TreeNode(3);
-        TreeNode node2 = new TreeNode(2);
-        TreeNode node1 = new TreeNode(1);
-        TreeNode noden3 = new TreeNode(-3);
-        TreeNode node11 = new TreeNode(11);
-        node10.left = node5;
-        node10.right = noden3;
-        node5.left = node3;
-        node5.right = node2;
-        node2.right = node1;
-        noden3.right = node11;
-        System.out.println(pathSum2(node10, 8));
-    }
-
 
     public static int pathSum(TreeNode root, int sum) { //该节点作为起点
         if (root == null) return 0;
@@ -48,10 +32,8 @@ public class lc437 {
     }
 
     public static int dfs(TreeNode root, int sum) { //一条路径向下走
-        if (root == null)
-            return 0;
-        if (root.val == sum)
-            return 1 + dfs(root.left, sum - root.val) + dfs(root.right, sum - root.val);//不要直接返回1，因为可能后边节点，或节点和为0
+        if (root == null) return 0;
+        if (root.val == sum) return 1 + dfs(root.left, sum - root.val) + dfs(root.right, sum - root.val);//不要直接返回1，因为可能后边节点，或节点和为0
         return dfs(root.left, sum - root.val) + dfs(root.right, sum - root.val);
     }
 
diff --git a/code/lc448.java b/code/lc448.java
index 802765b..364b76b 100644
--- a/code/lc448.java
+++ b/code/lc448.java
@@ -9,7 +9,7 @@ import java.util.List;
  * 难度：Easy
  * 分类：Array
  * 思路：把对应的数字放到对应的位置，最后遍历一遍，如果位置和数字不对应，则为缺失的值。
- * Tips：
+ * Tips：lc268 lc448 lc287
  */
 public class lc448 {
     public static void main(String[] args) {
diff --git a/code/lc494.java b/code/lc494.java
index 8bed557..26718d3 100644
--- a/code/lc494.java
+++ b/code/lc494.java
@@ -5,7 +5,10 @@ package code;
  * 难度：Medium
  * 分类：Dynamic Programming, Depth-first Search
  * 思路：可以用递归+mem的方法。也可以转化为0，1背包问题，注意dp的时候把下标移位。另一种方法是转化为子数组的和为(target + sum(nums))/2的问题，求解方法类似lc416
- * Tips：多抽象总结一下相关的问题，如何抽象出背包。对于这个数字，要么+，要么-，就两种情况。
+ * Tips：多抽象总结一下相关的问题，如何抽象出背包。对于这个数字，要么+，要么-，就两种情况。https://leetcode.com/problems/target-sum/discuss/97335/Short-Java-DP-Solution-with-Explanation
+ *       dp[i][j] 表示前i个元素和为j的方案个数
+ *       dp[i][j] = dp[i-1][j-nums[j]] + dp[i-1][j+nums[j]] //加减两种方案加起来
+ *       lc416, lc494
  */
 public class lc494 {
     public int findTargetSumWays(int[] nums, int S) {
diff --git a/code/lc538.java b/code/lc538.java
index e8a616d..2c9e782 100644
--- a/code/lc538.java
+++ b/code/lc538.java
@@ -17,14 +17,11 @@ public class lc538 {
     }
     int sum = 0;
     public TreeNode convertBST(TreeNode root) {
-        helper(root);
+        if(root==null) return null;
+        convertBST(root.right);
+        sum += root.val;
+        root.val = sum;
+        convertBST(root.left);
         return root;
     }
-    public void helper(TreeNode root){
-        if(root==null) return;
-        helper(root.right);
-        root.val = root.val + sum;
-        sum = root.val;
-        helper(root.left);
-    }
 }
diff --git a/code/lc543.java b/code/lc543.java
index 08c543b..6e77a3c 100644
--- a/code/lc543.java
+++ b/code/lc543.java
@@ -1,7 +1,7 @@
 package code;
 /*
  * 543. Diameter of Binary Tree
- * 题意：树种的最长路径
+ * 题意：树中的最长路径
  * 难度：Easy
  * 分类：Tree
  * 思路：和lc124思路一样，但lc124是Hard，这道竟然是Easy，哈哈哈
diff --git a/code/lc560.java b/code/lc560.java
index fd75474..f5eb844 100644
--- a/code/lc560.java
+++ b/code/lc560.java
@@ -9,6 +9,7 @@ import java.util.HashMap;
  * 分类：Array, Hash Table
  * 思路：求出累加和存在hashmap中，如果当前hashmap中存在sum-k，那么就是一个解
  * Tips：经典思路，记一下。lc437有类似思想。
+ *       lc303, lc437, lc560
  */
 public class lc560 {
     public int subarraySum(int[] nums, int k) {
diff --git a/code/lc617.java b/code/lc617.java
index 0927852..52913b2 100644
--- a/code/lc617.java
+++ b/code/lc617.java
@@ -42,13 +42,11 @@ public class lc617 {
     }
 
     public TreeNode mergeTrees2(TreeNode t1, TreeNode t2) {
-        if (t1 == null)
-            return t2;
-        if (t2 == null)
-            return t1;
-        t1.val += t2.val;
+        if(t1==null) return t2;     //这里注意一下，比较难想明白，可以记一下
+        if(t2==null) return t1;
         t1.left = mergeTrees(t1.left, t2.left);
         t1.right = mergeTrees(t1.right, t2.right);
+        t1.val += t2.val;
         return t1;
     }
 }
-- 
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