1. TwoSum
https://www.cnblogs.com/zhacai/p/10429120.html easy
2. Add Two Numbers
https://www.cnblogs.com/zhacai/p/10429155.html easy
15.3Sum
https://www.cnblogs.com/zhacai/p/10579514.html medium set
20.Valid Parentheses
https://www.cnblogs.com/zhacai/p/10429168.html easy
21.Merge Two Sorted Lists
https://www.cnblogs.com/zhacai/p/10429234.html easy
22.Generate Parentheses
https://www.cnblogs.com/zhacai/p/10599156.html medium DFS 剪枝
49.Group Anagrams
https://www.cnblogs.com/zhacai/p/10576638.html medium 哈希 map
53.Maximum Subarray
https://www.cnblogs.com/zhacai/p/10429247.html easy
70.Climbing Stairs
https://www.cnblogs.com/zhacai/p/10429253.html easy
72.Edit Distance
https://www.cnblogs.com/zhacai/p/10661858.html hard dp
79.Word Search
https://www.cnblogs.com/zhacai/p/10641454.html medium DFS
94. Binary Tree Inorder Traversal
https://www.cnblogs.com/zhacai/p/10435799.html medium 树的中序遍历<递归><迭代（栈）>
98.Validate Binary Search Tree
https://www.cnblogs.com/zhacai/p/10592337.html medium 二叉搜索树 中序遍历 递归
102.Binary Tree Level Order Traversal
https://www.cnblogs.com/zhacai/p/10598674.html medium 树 BFS
104.Maxinum Depth of Binary Tree
https://www.cnblogs.com/zhacai/p/10429258.html easy 树DFS
121. Best Time to Buy and Sell Stock
https://www.cnblogs.com/zhacai/p/10429264.html easy
136. Single Number
https://www.cnblogs.com/zhacai/p/10429270.html easy
141. Linked List Cycle
https://www.cnblogs.com/zhacai/p/10560803.html easy 链表 快慢指针
142. Linked List Cycle II
https://www.cnblogs.com/zhacai/p/10561152.html medium 链表 快慢指针
146. LRU Cache
https://www.cnblogs.com/zhacai/p/10665121.html hard LRUCache linkedHashMap hashmap+双向链表
152. Maximum Product Subarray
https://www.cnblogs.com/zhacai/p/10644095.html medium dp
155. Min Stack
https://www.cnblogs.com/zhacai/p/10429280.html easy
169.Majority Element
https://www.cnblogs.com/zhacai/p/10429286.html easy
200.Number of Islands
https://www.cnblogs.com/zhacai/p/10662241.html medium dfs 并查集
206.ReverseLinked List
https://www.cnblogs.com/zhacai/p/10429295.html easy 链表
208.Implement Trie(Prefix Tree)
https://www.cnblogs.com/zhacai/p/10640769.html medium 字典树
226.Invert Binary Tree
https://www.cnblogs.com/zhacai/p/10431018.html easy 树 递归
236.Lowest Common Ancestor of a Binary Tree
https://www.cnblogs.com/zhacai/p/10592779.html medium 树 递归
239.Sliding Window Maximum
https://www.cnblogs.com/zhacai/p/10567783.html hard 双向队列 大顶堆
283.Move Zeroes
https://www.cnblogs.com/zhacai/p/10431000.html easy
300.Longest Increasing Subsequence
https://www.cnblogs.com/zhacai/p/10661330.html medium dp
309.Best Time to Buy and Sell Stock with Cooldown
https://www.cnblogs.com/zhacai/p/10655970.html medium dp
322.Coin Change
https://www.cnblogs.com/zhacai/p/10661489.html medium dp
338. Counting Bits
https://www.cnblogs.com/zhacai/p/10430986.html easy 位运算 找规律
347. Top K Frequent Elements(待完善)
https://www.cnblogs.com/zhacai/p/10436261.html medium 排序 map
438.Find All Anagrams in a String
https://www.cnblogs.com/zhacai/p/10596274.html easy 滑动窗口 map
448. Find All Numbers Disappeared in an Array
Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements of [1, n] inclusive that do not appear in this array. Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.
Example:
Input:
[4,3,2,7,8,2,3,1] Output:
[5,6]
public List<Integer> findDisappearedNumbers(int[] nums) {
for (int i = 0; i < nums.length; i++) {
int x = nums[i]<0?-nums[i]:nums[i];
if(nums[x-1]>0){
nums[x-1]= -nums[x-1];
}
}
List<Integer> re = new ArrayList<Integer>();
for (int i = 0; i < nums.length; i++) {
if(nums[i]>0){
re.add(i+1);
}
}
return re;
}
461.Hamming Distance
The Hamming distance between two integers is the number of positions at which the corresponding bits are different.Given two integers x and y, calculate the Hamming distance.
Note:0 ≤ x, y < 231.
Example:
Input: x = 1, y = 4 Output: 2 Explanation:
1 (0 0 0 1)
4 (0 1 0 0)
↑ ↑ The above arrows point to positions where the corresponding bits are different.
public int hammingDistance(int x, int y) {//位运算
int n = x^y;
int re = 0;
//计算整数二进制中1的个数
while(0!=n){
n=n&(n-1);
re++;
}
return re;
}
538.Convert BST to Greater Tree
Given a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus sum of all keys greater than the original key in BST.
Example:
Input: The root of a Binary Search Tree like this:
5
/ \
2 13 Output: The root of a Greater Tree like this:
18
/ \
20 13
class Solution {
int sum=0;
public TreeNode convertBST(TreeNode root) {//树
if(null!=root){
TreeNode right = convertBST(root.right);
root.val+=sum;
sum = root.val;
TreeNode left = convertBST(root.left);
}
return root;
}
}
543.Diameter of Binary Tree
Given a binary tree, you need to compute the length of the diameter of the tree. The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root.
Example:
Given a binary tree
1
/ \
2 3
/ \
4 5
Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].
Note: The length of path between two nodes is represented by the number of edges between them.
class Solution {
int re;
public int diameterOfBinaryTree(TreeNode root) {//树的深度 递归
re =0;
maxDepth(root);
return re;
}
public int maxDepth(TreeNode root) {
if(null==root){
return 0;
}
int ld=maxDepth(root.left);
int lr = maxDepth(root.right);
int d = ld>lr?(ld+1):(lr+1);
re =re>(ld+lr)?re:(ld+lr);
return d;
}
}
617.Merge Two Binary Trees
Given two binary trees and imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not. You need to merge them into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of new tree.
Example 1:
Input:
Tree 1 Tree 2
1 2
/ \ / \
3 2 1 3
/ \ \
5 4 7
Output:
Merged tree:
3
/ \
4 5
/ \ \
5 4 7
Note: The merging process must start from the root nodes of both trees.
public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {//树 递归
if(null==t1&&null==t2){
return null;
}
if(null==t1){
return t2;
}
if(null==t2){
return t1;
}
TreeNode left=mergeTrees(t1.left,t2.left);
TreeNode right = mergeTrees(t1.right,t2.right);
TreeNode node = new TreeNode(t1.val+t2.val);
node.left=left;
node.right=right;
return node;
}
简洁写法
public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
if (t1 == null)
return t2;
if (t2 == null)
return t1;
t1.val += t2.val;
t1.left = mergeTrees(t1.left, t2.left);
t1.right = mergeTrees(t1.right, t2.right);
return t1;
}
非递归方法
public class Solution {
public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
if (t1 == null)
return t2;
Stack < TreeNode[] > stack = new Stack < > ();
stack.push(new TreeNode[] {t1, t2});
while (!stack.isEmpty()) {
TreeNode[] t = stack.pop();
if (t[0] == null || t[1] == null) {
continue;
}
t[0].val += t[1].val;
if (t[0].left == null) {
t[0].left = t[1].left;
} else {
stack.push(new TreeNode[] {t[0].left, t[1].left});
}
if (t[0].right == null) {
t[0].right = t[1].right;
} else {
stack.push(new TreeNode[] {t[0].right, t[1].right});
}
}
return t1;
}
}
771. Jewels and Stones
https://www.cnblogs.com/zhacai/p/10429670.html easy
7.Reverse Integer
Given a 32-bit signed integer, reverse digits of an integer.
Example 1:
Input: 123
Output: 321
Example 2:
Input: -123
Output: -321
Example 3:
Input: 120
Output: 21
Note:Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
public int reverse(int x) {
String str= String.valueOf(x);
StringBuffer strb;
int re=0;
if('-'==str.charAt(0)){
strb = new StringBuffer(str.substring(1,str.length())).reverse();
strb.insert(0,"-");
}
else {
strb= new StringBuffer(str).reverse();
}
try{
re =Integer.valueOf(strb.toString());
}catch (Exception e){
}
return re;
}
使用数学方法的优解
public int reverse(int x) {
int re = 0;
while(0!=x){
int temp = re * 10 + x % 10;
if(temp / 10 != re)//如果整数不溢出显然相等
return 0;
re = temp;
x /= 10;
}
return re;
}