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Commit 2d9f477

Browse files
committed
20190320
1 parent de13f7e commit 2d9f477

22 files changed

+116
-53
lines changed

‎code/lc112.java

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package code;
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/*
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* 112. Path Sum
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* 题意:是否存在跟到叶子节点的和为sum
5+
* 难度:Easy
6+
* 分类:
7+
* 思路:
8+
* Tips:lc112, lc113, lc437, lc129, lc124, lc337
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* 总结一下 lc112 找跟到叶子和为sum的路径是否存在
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* lc113 把lc112的路径找出来
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* lc437 是任意一条向下走的路径
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* lc129 计算所有的和
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* lc124 任意一条路径
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* lc337 树的dp
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*/
16+
public class lc112 {
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public class TreeNode {
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int val;
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TreeNode left;
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TreeNode right;
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TreeNode(int x) {
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val = x;
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}
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}
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public boolean hasPathSum(TreeNode root, int sum) {
26+
if(root==null) return false;
27+
if(root.val==sum&&root.left==null&&root.right==null) return true;
28+
return hasPathSum(root.left, sum-root.val)||hasPathSum(root.right, sum-root.val);
29+
}
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}

‎code/lc113.java

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -8,7 +8,7 @@
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* 难度:Medium
99
* 分类:Tree, Depth-first Search
1010
* 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
11-
* Tips:lc124
11+
* Tips:lc112, lc113, lc437, lc129, lc124, lc337
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*/
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public class lc113 {
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public class TreeNode {

‎code/lc124.java

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -6,7 +6,7 @@
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* 分类:Tree, Depth-first Search
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* 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。
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* 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
9-
* lc133
9+
* lc112, lc113, lc437, lc129, lc124, lc337, lc543
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*/
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public class lc124 {
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public class TreeNode {

‎code/lc129.java

Lines changed: 28 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,28 @@
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package code;
2+
/*
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* 129. Sum Root to Leaf Numbers
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* 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
5+
* 难度:Medium
6+
* 分类:Tree, Depth-first Search
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* 思路:dfs
8+
* Tips:lc112, lc113, lc437, lc129, lc124, lc337
9+
*/
10+
public class lc129 {
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public class TreeNode {
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int val;
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TreeNode left;
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TreeNode right;
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TreeNode(int x) {
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val = x;
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}
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}
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public int sumNumbers(TreeNode root) {
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if(root==null) return 0;
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return helper(root, 0);
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}
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public int helper(TreeNode root, int sum){
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if(root==null) return 0; //一条路径,另一边返回0
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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
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return helper(root.left, sum*10+root.val) + helper(root.right, sum*10+root.val);
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}
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}

‎code/lc151.java

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package code;
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import java.util.Arrays;
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import java.util.Collections;
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/*
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* 151. Reverse Words in a String
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* 题意:反转字符串中的单词
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* 难度:Medium
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* 分类:String
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* 思路:难点在中间的空格可能是多个空格,注意如何解决
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* Tips:
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*/
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public class lc151 {
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public String reverseWords(String s) {
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String[] words = s.trim().split(" +"); //+号匹配多个
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Collections.reverse(Arrays.asList(words));
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return String.join(" ", words);
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}
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}

‎code/lc236.java

Lines changed: 3 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -35,7 +35,7 @@ public TreeNode lowestCommonAncestor2(TreeNode root, TreeNode p, TreeNode q) {
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parent.put(root, null);
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stack.push(root);
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38-
while (!parent.containsKey(p) || !parent.containsKey(q)) {
38+
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
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TreeNode node = stack.pop();
4040
if (node.left != null) {
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parent.put(node.left, node);
@@ -47,11 +47,11 @@ public TreeNode lowestCommonAncestor2(TreeNode root, TreeNode p, TreeNode q) {
4747
}
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}
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Set<TreeNode> ancestors = new HashSet<>();
50-
while (p != null) {
50+
while (p != null) {//p的路径节点添加到hashset中
5151
ancestors.add(p);
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p = parent.get(p);
5353
}
54-
while (!ancestors.contains(q))
54+
while (!ancestors.contains(q))//第一个hashset中遇到的节点,就是最近公共祖先
5555
q = parent.get(q);
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return q;
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}

‎code/lc239.java

Lines changed: 3 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -1,10 +1,10 @@
11
package code;
22
/*
33
* 239. Sliding Window Maximum
4-
* 题意:滑动窗口最大值
4+
* 题意:滑动窗口中最大值
55
* 难度:Hard
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* 分类:Heap
7-
* 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。
7+
* 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。
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* Tips:与lc84做比较,84是递增栈
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*/
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import java.util.ArrayDeque;
@@ -25,7 +25,7 @@ public static int[] maxSlidingWindow(int[] nums, int k) {
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return new int[]{};
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int[] res = new int[nums.length-k+1];
2727
int cur = 0;
28-
Deque<Integer> dq = new ArrayDeque();
28+
Deque<Integer> dq = new ArrayDeque();//队列里是递减的
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for (int i = 0; i < nums.length ; i++) {
3030
if( !dq.isEmpty() && dq.peekFirst()<=i-k)
3131
dq.removeFirst();

‎code/lc268.java

Lines changed: 2 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -7,7 +7,7 @@
77
* 思路:两种巧妙的方法,时间空间都是O(1)
88
* 异或
99
* 求和以后,减去所有
10-
* Tips:
10+
* Tips:lc268 lc448 lc287
1111
*/
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public class lc268 {
1313
public int missingNumber(int[] nums) {
@@ -16,7 +16,7 @@ public int missingNumber(int[] nums) {
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return res;
1717
}
1818
public int missingNumber2(int[] nums) {
19-
int res = nums.length;
19+
int res = nums.length;//异或上长度
2020
for(int i=0; i<nums.length; i++) res^=i^nums[i];
2121
return res;
2222
}

‎code/lc279.java

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -17,7 +17,7 @@ public static int numSquares(int n) {
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int[] dp = new int[n];
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Arrays.fill(dp,Integer.MAX_VALUE);
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for (int i = 1; i <= n ; i++) { //两个for循环
20-
for (int j=1; j<=i ; j++) {
20+
for (int j=1; j*j<=i ; j++) {
2121
if(j*j==i)
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dp[i-1] = 1;
2323
if(j*j<i){

‎code/lc287.java

Lines changed: 2 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -3,11 +3,12 @@
33
* 287. Find the Duplicate Number
44
* 题意:n+1个数属于[1~n],找出重复的那个数
55
* 难度:Medium
6-
* 分类:Array, Two Pointers, Binary Search
6+
* 分类:Array, Two Pointers, Binary Searn+1个数属于[1~n],找出重复的那个数ch
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* 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。
88
* 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
99
* 二分查找,每次看一边数字的个数, O(nlog(n)) O(1)
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* Tips:剑指offer原题
11+
* lc268 lc448 lc287
1112
*/
1213
public class lc287 {
1314
public int findDuplicate(int[] nums) { //该方法修改了数组,是错误的,没看清题意

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