3，1，42，200，15
121，128，49，25，88
5，146，70，2，4
21，33，55，27，560
11，20，31，53，236
300，26，215，279，438
135，148，9，169，76
22，101，14，54，56
72，206，152，80，39
46，62，104，122，179
3. Longest Substring Without Repeating Characters
Medium
Given a string s, find the length of the longest substring without repeating characters.
Example 1:
Input: s = "abcabcbb" Output: 3 Explanation: The answer is "abc", with the length of 3.
class Solution {
public int lengthOfLongestSubstring(String s) {
Set<Character> set = new HashSet<>();
char[] ch = s.toCharArray();
int left = 0;
int count = 0;
int ans = 0;
for(int right = 0; right < ch.length; right++){
if(!set.contains(ch[right])){
set.add(ch[right]);
count = count + 1;
ans = Math.max(ans, count);
}else{
while(set.contains(ch[right])){
set.remove(ch[left]);
left++;
count--;
}
set.add(ch[right]);
count++;
}
}
return ans;
}
}class Solution {
public int lengthOfLongestSubstring(String s) {
int[] ch = new int[128];
for(int i=0;i<128;i++){
ch[i]=-1;
}
int ans=0;
int st=0;
for(int i=0;i<s.length();i++){
int c = s.charAt(i);
if(ch[c]!=-1){
st=Math.max(st,ch[c]);
}
ans=Math.max(ans,i-st+1);
ch[c]=i+1;
}
return ans;
}
}Easy
Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
You can return the answer in any order.
class Solution {
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> hashtable = new HashMap<Integer, Integer>();
for (int i = 0; i < nums.length; ++i) {
if (hashtable.containsKey(target - nums[i])) {
return new int[]{hashtable.get(target - nums[i]), i};
}
hashtable.put(nums[i], i);
}
return new int[0];
}
}Hard
Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it can trap after raining.
class Solution {
public int trap(int[] height) {
int ans = 0;
int left = 0, right = height.length - 1;
int leftMax = 0, rightMax = 0;
while (left < right) {
leftMax = Math.max(leftMax, height[left]);
rightMax = Math.max(rightMax, height[right]);
if (height[left] < height[right]) {
ans += leftMax - height[left];
++left;
} else {
ans += rightMax - height[right];
--right;
}
}
return ans;
}
}Medium
Given an m x n 2D binary grid grid which represents a map of '1's (land) and '0's (water), return the number of islands.
An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.
class Solution {
void dfs(char[][] grid, int r, int c) {
int nr = grid.length;
int nc = grid[0].length;
if (r < 0 || c < 0 || r >= nr || c >= nc || grid[r][c] == '0') {
return;
}
grid[r][c] = '0';
dfs(grid, r - 1, c);
dfs(grid, r + 1, c);
dfs(grid, r, c - 1);
dfs(grid, r, c + 1);
}
public int numIslands(char[][] grid) {
if (grid == null || grid.length == 0) {
return 0;
}
int nr = grid.length;
int nc = grid[0].length;
int num_islands = 0;
for (int r = 0; r < nr; ++r) {
for (int c = 0; c < nc; ++c) {
if (grid[r][c] == '1') {
++num_islands;
dfs(grid, r, c);
}
}
}
return num_islands;
}
}Medium
Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.
Notice that the solution set must not contain duplicate triplets.
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
List<List<Integer>> ans = new ArrayList<>();
Arrays.sort(nums);
for(int i = 0; i < nums.length-2; i++){
if(nums[i] + nums[i + 1] + nums[i + 2] > 0){
break;
}
if(nums[i] + nums[nums.length - 1] + nums[nums.length -2] < 0){
continue;
}
if(i > 0 && nums[i] == nums[i - 1]){
continue;
}
int j = i + 1;
int k = nums.length - 1;
while(j < k){
if(nums[i] + nums[j] + nums[k] > 0){
k--;
}else if(nums[i] + nums[j] + nums[k] < 0){
j++;
}else{
List<Integer> list = new ArrayList<>();
list.add(nums[i]);
list.add(nums[j]);
list.add(nums[k]);
ans.add(list);
j++;
k--;
while(j < k && nums[j] == nums[j -1]){
j++;
}
while(j < k && nums[k] == nums[k + 1]){
k--;
}
}
}
}
return ans;
}
}121. Best Time to Buy and Sell Stock
Easy
You are given an array prices where prices[i] is the price of a given stock on the ith day.
You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock.
Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.
Example 1:
Input: prices = [7,1,5,3,6,4] Output: 5 Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5. Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.
class Solution {
public int maxProfit(int[] prices) {
if (prices.length == 1) {
return 0;
}
int[][] dp = new int[prices.length][2];//dp[i][0]表示第i天持有股票的最大收益，dp[i][1]表示第i天不持有股票的最大收益
dp[0][0] = -prices[0];
dp[0][1] = 0;
for (int i = 1; i < prices.length; i++) {
dp[i][0] = Math.max(dp[i - 1][0], -prices[i]);//分为是不是今天买入
dp[i][1] = Math.max(dp[i - 1][1], dp[i - 1][0] + prices[i]);//分为是不是今天买
}
return dp[prices.length - 1][1];
}
}class Solution {
public int maxProfit(int[] prices) {
int cost = Integer.MAX_VALUE, profit = 0;
for (int price : prices) {
cost = Math.min(cost, price);
profit = Math.max(profit, price - cost);
}
return profit;
}
}128. Longest Consecutive Sequence
Medium
Given an unsorted array of integers nums, return the length of the longest consecutive elements sequence.
You must write an algorithm that runs in O(n) time.
Example 1:
Input: nums = [100,4,200,1,3,2]
Output: 4
Explanation: The longest consecutive elements sequence is [1, 2, 3, 4]. Therefore its length is 4.
class Solution {
public int longestConsecutive(int[] nums) {
Set<Integer> numSet = new HashSet<Integer>();
for (int num : nums) {
numSet.add(num);
}
int longestLength = 0;
for (int num : numSet) {
if (!numSet.contains(num - 1)) {
int currentNum = num;
int currentLength = 1;
while (numSet.contains(currentNum + 1)) {
currentNum += 1;
currentLength += 1;
}
longestLength = Math.max(longestLength, currentLength);
}
}
return longestLength;
}
}Solved
Medium
Topics
Companies
Given an array of strings strs, group the anagrams together. You can return the answer in any order.
An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.
class Solution {
public List<List<String>> groupAnagrams(String[] strs) {
Map<String, List<String>> map = new HashMap<String, List<String>>();
for (String str : strs) {
char[] array = str.toCharArray();
Arrays.sort(array);
String key = new String(array);
List<String> list = map.getOrDefault(key, new ArrayList<String>());
list.add(str);
map.put(key, list);
}
return new ArrayList<List<String>>(map.values());
}
}class Solution {
public List<List<String>> groupAnagrams(String[] strs) {
Map<String, List<String>> map = new HashMap<String, List<String>>();
for (String str : strs) {
int[] counts = new int[26];
int length = str.length();
for (int i = 0; i < length; i++) {
counts[str.charAt(i) - 'a']++;
}
// 将每个出现次数大于 0 的字母和出现次数按顺序拼接成字符串，作为哈希表的键
StringBuffer sb = new StringBuffer();
for (int i = 0; i < 26; i++) {
if (counts[i] != 0) {
sb.append((char) ('a' + i));
sb.append(counts[i]);
}
}
String key = sb.toString();
List<String> list = map.getOrDefault(key, new ArrayList<String>());
list.add(str);
map.put(key, list);
}
return new ArrayList<List<String>>(map.values());
}
}
Hard
Topics
Companies
Given the head of a linked list, reverse the nodes of the list k at a time, and return the modified list.
k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes, in the end, should remain as it is.
You may not alter the values in the list's nodes, only nodes themselves may be changed.
class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
ListNode hair = new ListNode(0);
hair.next = head;
ListNode pre = hair;
while (head != null) {
ListNode tail = pre;
// 查看剩余部分长度是否大于等于 k
for (int i = 0; i < k; ++i) {
tail = tail.next;
if (tail == null) {
return hair.next;
}
}
ListNode nex = tail.next;
ListNode[] reverse = myReverse(head, tail);
head = reverse[0];
tail = reverse[1];
// 把子链表重新接回原链表
pre.next = head;
tail.next = nex;
pre = tail;
head = tail.next;
}
return hair.next;
}
public ListNode[] myReverse(ListNode head, ListNode tail) {
ListNode prev = tail.next;
ListNode p = head;
while (prev != tail) {
ListNode nex = p.next;
p.next = prev;
prev = p;
p = nex;
}
return new ListNode[]{tail, head};
}
}
Easy
You are given two integer arrays nums1 and nums2, sorted in non-decreasing order, and two integers m and n, representing the number of elements in nums1 and nums2 respectively.
Merge nums1 and nums2 into a single array sorted in non-decreasing order.
The final sorted array should not be returned by the function, but instead be stored inside the array nums1. To accommodate this, nums1 has a length of m + n, where the first m elements denote the elements that should be merged, and the last n elements are set to 0 and should be ignored. nums2 has a length of n.
Example 1:
Input: nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3 Output: [1,2,2,3,5,6] Explanation: The arrays we are merging are [1,2,3] and [2,5,6]. The result of the merge is [1,2,2,3,5,6] with the underlined elements coming from nums1.
class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
int p1 = 0, p2 = 0;
int[] sorted = new int[m + n];
int cur;
while (p1 < m || p2 < n) {
if (p1 == m) {
cur = nums2[p2++];
} else if (p2 == n) {
cur = nums1[p1++];
} else if (nums1[p1] < nums2[p2]) {
cur = nums1[p1++];
} else {
cur = nums2[p2++];
}
sorted[p1 + p2 - 1] = cur;
}
for (int i = 0; i != m + n; ++i) {
nums1[i] = sorted[i];
}
}
}
5. Longest Palindromic Substring
Medium
Given a string s, return the longest
palindromic substring in s.
class Solution {
public String longestPalindrome(String s) {
if (s == null || s.length() < 1) {
return "";
}
int start = 0, end = 0;
for (int i = 0; i < s.length(); i++) {
int len1 = expandAroundCenter(s, i, i);
int len2 = expandAroundCenter(s, i, i + 1);
int len = Math.max(len1, len2);
if (len > end - start) {
start = i - (len - 1) / 2;
end = i + len / 2;
}
}
return s.substring(start, end + 1);
}
public int expandAroundCenter(String s, int left, int right) {
while (left >= 0 && right < s.length() && s.charAt(left) == s.charAt(right)) {
--left;
++right;
}
return right - left - 1;
}
}Medium
Topics
Companies
Design a data structure that follows the constraints of a Least Recently Used (LRU) cache.
Implement the LRUCache class:
LRUCache(int capacity)Initialize the LRU cache with positive sizecapacity.int get(int key)Return the value of thekeyif the key exists, otherwise return-1.void put(int key, int value)Update the value of thekeyif thekeyexists. Otherwise, add thekey-valuepair to the cache. If the number of keys exceeds thecapacityfrom this operation, evict the least recently used key.
The functions get and put must each run in O(1) average time complexity.
class LRUCache {
class Node{
int key;
int value;
Node prev;
Node next;
Node(int key, int value){
this.key= key;
this.value= value;
}
}
public Node[] map;
public int count, capacity;
public Node head, tail;
public LRUCache(int capacity) {
this.capacity= capacity;
count= 0;
map= new Node[10_000+1];
head= new Node(0,0);
tail= new Node(0,0);
head.next= tail;
tail.prev= head;
head.prev= null;
tail.next= null;
}
public void deleteNode(Node node){
node.prev.next= node.next;
node.next.prev= node.prev;
return;
}
public void addToHead(Node node){
node.next= head.next;
node.next.prev= node;
node.prev= head;
head.next= node;
return;
}
public int get(int key) {
if( map[key] != null ){
Node node= map[key];
int nodeVal= node.value;
deleteNode(node);
addToHead(node);
return nodeVal;
}
else
return -1;
}
public void put(int key, int value) {
if(map[key] != null){
Node node= map[key];
node.value= value;
deleteNode(node);
addToHead(node);
} else {
Node node= new Node(key,value);
map[key]= node;
if(count < capacity){
count++;
addToHead(node);
}
else {
map[tail.prev.key]= null;
deleteNode(tail.prev);
addToHead(node);
}
}
return;
}
}public class LRUCache {
class DLinkedNode {
int key;
int value;
DLinkedNode prev;
DLinkedNode next;
public DLinkedNode() {}
public DLinkedNode(int _key, int _value) {key = _key; value = _value;}
}
private Map<Integer, DLinkedNode> cache = new HashMap<Integer, DLinkedNode>();
private int size;
private int capacity;
private DLinkedNode head, tail;
public LRUCache(int capacity) {
this.size = 0;
this.capacity = capacity;
// 使用伪头部和伪尾部节点
head = new DLinkedNode();
tail = new DLinkedNode();
head.next = tail;
tail.prev = head;
}
public int get(int key) {
DLinkedNode node = cache.get(key);
if (node == null) {
return -1;
}
// 如果 key 存在，先通过哈希表定位，再移到头部
moveToHead(node);
return node.value;
}
public void put(int key, int value) {
DLinkedNode node = cache.get(key);
if (node == null) {
// 如果 key 不存在，创建一个新的节点
DLinkedNode newNode = new DLinkedNode(key, value);
// 添加进哈希表
cache.put(key, newNode);
// 添加至双向链表的头部
addToHead(newNode);
++size;
if (size > capacity) {
// 如果超出容量，删除双向链表的尾部节点
DLinkedNode tail = removeTail();
// 删除哈希表中对应的项
cache.remove(tail.key);
--size;
}
}
else {
// 如果 key 存在，先通过哈希表定位，再修改 value，并移到头部
node.value = value;
moveToHead(node);
}
}
private void addToHead(DLinkedNode node) {
node.prev = head;
node.next = head.next;
head.next.prev = node;
head.next = node;
}
private void removeNode(DLinkedNode node) {
node.prev.next = node.next;
node.next.prev = node.prev;
}
private void moveToHead(DLinkedNode node) {
removeNode(node);
addToHead(node);
}
private DLinkedNode removeTail() {
DLinkedNode res = tail.prev;
removeNode(res);
return res;
}
}
Medium
Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
class Solution {
public List<String> generateParenthesis(int n) {
List<String> ans = new ArrayList<String>();
backtrack(ans, new StringBuilder(), 0, 0, n);
return ans;
}
public void backtrack(List<String> ans, StringBuilder cur, int open, int close, int max) {
if (cur.length() == max * 2) {
ans.add(cur.toString());
return;
}
if (open < max) {
cur.append('(');
backtrack(ans, cur, open + 1, close, max);
cur.deleteCharAt(cur.length() - 1);
}
if (close < open) {
cur.append(')');
backtrack(ans, cur, open, close + 1, max);
cur.deleteCharAt(cur.length() - 1);
}
}
}Easy
Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).
class Solution {
public boolean isSymmetric(TreeNode root) {
return compare(root.left, root.right);
}
private boolean compare(TreeNode left, TreeNode right) {
if (left == null && right == null) {
return true;
}
if (left == null || right == null) {
return false;
}
if (left.val != right.val) {
return false;
}
return compare(left.right, right.left) && compare(left.left, right.right);
}
}Easy
Write a function to find the longest common prefix string amongst an array of strings.
If there is no common prefix, return an empty string "".
class Solution {
public String longestCommonPrefix(String[] strs) {
if (strs == null || strs.length == 0) {
return "";
}
String prefix = strs[0];
int count = strs.length;
for (int i = 1; i < count; i++) {
prefix = longestCommonPrefix(prefix, strs[i]);
if (prefix.length() == 0) {
break;
}
}
return prefix;
}
public String longestCommonPrefix(String str1, String str2) {
int length = Math.min(str1.length(), str2.length());
int index = 0;
while (index < length && str1.charAt(index) == str2.charAt(index)) {
index++;
}
return str1.substring(0, index);
}
}