一、数组
8） 双指针 ---- 滑动窗口
例题：
3. Longest Substring Without Repeating Characters
描述：Given a string, find the length of the longest substring without repeating characters.
题解：时间：92.67%，空间：87.02%
public int lengthOfLongestSubstring(String s) {
// check
if(s == null || s.length() == 0) return 0;
// initial
int[] freq = new int[256];
Arrays.fill(freq, -1);
int l = 0, r = 0, res = 0;
while(r < s.length()){
if(freq[s.charAt(r)] != -1) l = Math.max(l, freq[s.charAt(r)] + 1);
freq[s.charAt(r)] = r++;
res = Math.max(res, r - l);
}
return res;
}
练习：
438. Find All Anagrams in a String
76. Minimum Window Substring
二、查找问题
1）查找有无：是否存在
2）查找对应关系：出现了几次
349. Intersection of Two Arrays // 熟悉Set
描述：Given two arrays, write a function to compute their intersection.
Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2]
public int[] intersection(int[] nums1, int[] nums2) {
if(nums1 == null || nums1.length == 0 || nums2 == null || nums2.length == 0)
return new int[0];
Set<Integer> set1 = new HashSet<Integer>();
Set<Integer> setTemp = new HashSet<Integer>();
for(int num : nums1) set1.add(num);
for(int num : nums2) if(set1.contains(num)) setTemp.add(num);
int k = setTemp.size();
int[] res = new int[k];
for(int num : setTemp) res[--k] = num;
return res;
}
350. Intersection of Two Arrays II // 熟悉 Map
描述：Given two arrays, write a function to compute their intersection.
Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2,2]
public int[] intersect(int[] nums1, int[] nums2) {
if(nums1 == null || nums1.length == 0 || nums2 == null || nums2.length == 0)
return new int[0];
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
for(int num : nums1){
map.put(num, map.getOrDefault(num, 0) + 1);
}
List<Integer> list = new ArrayList<Integer>();
for(int num : nums2){
if(map.get(num) != null && map.get(num) > 0){
list.add(num);
map.put(num, map.get(num) - 1);
}
}
int k = list.size();
int[] arr = new int[k];
for(int num : list) arr[--k] = num;
return arr;
}
242. Valid Anagram
题目描述：Given two strings s and t , write a function to determine if t is an anagram of s.
202. Happy Number
290. Word Pattern
205. Isomorphic Strings
451. Sort Characters By Frequency
查找表的经典问题：
1. Two Sum
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
时间：99.77%，空间：88.15%
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
for(int i = 0; i < nums.length; i++){
int de = target - nums[i];
if(map.get(de) != null) return new int[]{map.get(de), i};
map.put(nums[i], i);
}
return new int[2];
}
15. 3Sum
18. 4Sum
16. 3Sum Closest
454. 4Sum II
描述：Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero.
To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of -2^28 to 2^28 - 1 and the result is guaranteed to be at most 2^31 - 1.
public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
for(int a : A)
for(int b : B)
map.put(a + b, map.getOrDefault(a + b, 0) + 1);
int res = 0;
for(int c : C)
for(int d : D){
if(map.getOrDefault(0 - c - d, 0) > 0){
res += map.get(0 - c - d);
}
}
return res;
}
49. Group Anagrams
447. Number of Boomerangs
Given n points in the plane that are all pairwise distinct, a "boomerang" is a tuple of points (i, j, k) such that the distance between i and j equals the distance between i and k (the order of the tuple matters).
Find the number of boomerangs. You may assume that nwill be at most 500 and coordinates of points are all in the range [-10000, 10000] (inclusive).
public int numberOfBoomerangs(int[][] points) {
if(points == null || points.length == 0) return 0;
Map<Integer, Integer> maps = new HashMap<Integer, Integer>();
int res = 0;
for(int i = 0; i < points.length; i++){
for(int j = 0; j < points.length; j++){
int distance = (points[i][0] - points[j][0]) * (points[i][0] - points[j][0]) +
(points[i][1] - points[j][1]) * (points[i][1] - points[j][1]);
maps.put(distance, maps.getOrDefault(distance, 0) + 1);
}
for(Map.Entry<Integer, Integer> map : maps.entrySet())
res += map.getValue() * (map.getValue() - 1);
maps.clear();
}
return res;
}
149. Max Points on a Line
滑动窗口+查找表
219. Contains Duplicate II
Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the absolute difference between i and j is at most k.
public boolean containsNearbyDuplicate(int[] nums, int k) {
if(nums == null || nums.length == 0 || k <= 0) return false;
Set<Integer> set = new HashSet<Integer>();
for(int i = 0; i < nums.length; i++){
if(i < k + 1) {
set.add(nums[i]);
if(set.size() < i+1) return true;
continue;
}
set.remove(nums[i - k - 1]);
set.add(nums[i]);
if(set.size() < k+1) return true;
}
return false;
}
217. Contains Duplicate
220. Contains Duplicate III
Given an array of integers, find out whether there are two distinct indices i and j in the array such that the absolute difference between nums[i] and nums[j] is at most t and the absolute difference between iand j is at most k.
不是优秀的解法：
public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
if(nums == null || nums.length == 0 || t < 0 || k <= 0) return false;
TreeSet<Long> set = new TreeSet<Long>();
for(int i = 0; i < nums.length; i++){
if(set.ceiling((long)nums[i] - t) != null && set.floor((long)nums[i] + t) != null
&& set.ceiling((long)nums[i] - t) - set.floor((long)nums[i] + t) <= 2 * (long)t)
return true;
set.add((long)nums[i]);
if(set.size() > k) set.remove((long)nums[i - k]);
}
return false;
}
链表
206. Reverse Linked List
Reverse a singly linked list.
Input: 1->2->3->4->5->NULL
Output: 5->4->3->2->1->NULL
public ListNode reverseList(ListNode head) {
ListNode pre = null;
ListNode cur = head;
ListNode next = null;
while(cur != null){
next = cur.next;
cur.next = pre;
pre = cur;
cur = next;
}
return pre;
}
92. Reverse Linked List II
链表基础操作
83. Remove Duplicates from Sorted List
86. Partition List
328. Odd Even Linked List
2. Add Two Numbers
445. Add Two Numbers II
虚拟头节点
203. Remove Linked List Elements
Remove all elements from a linked list of integers that have value val.
public ListNode removeElements(ListNode head, int val) {
ListNode tempNode = new ListNode(0);
tempNode.next = head;
ListNode cur = tempNode;
while(cur.next != null){
if(cur.next.val == val) cur.next = cur.next.next;
else cur = cur.next;
}
return tempNode.next;
}
82. Remove Duplicates from Sorted List II
21. Merge Two Sorted Lists
24. Swap Nodes in Pairs
Given a linked list, swap every two adjacent nodes and return its head.
You may not modify the values in the list's nodes, only nodes itself may be changed.
public ListNode swapPairs(ListNode head) {
ListNode top = new ListNode(0);
top.next = head;
ListNode node = top;
while(node.next != null && node.next.next != null){
ListNode first = node.next;
ListNode second = first.next;
node.next = second;
first.next = second.next;
second.next =first;
node = node.next.next;
}
return top.next;
}
25. Reverse Nodes in k-Group
链表排序
147. Insertion Sort List
148. Sort List
改变值
237. Delete Node in a Linked List
class Solution {
public void deleteNode(ListNode node) {
node.val = node.next.val;
node.next = node.next.next;
}
}
链表与双指针
19. Remove Nth Node From End of List
Given a linked list, remove the n-th node from the end of list and return its head.
public ListNode removeNthFromEnd(ListNode head, int n) {
// LeetCode对特殊情况出现的少，比如 head 为空， n的验证等
ListNode top = new ListNode(0);
top.next = head;
ListNode l = top, r = top;
while(n > 0){
r = r.next;
n--;
}
while(r.next != null){
l = l.next;
r = r.next;
}
l.next = l.next.next;
return top.next;
}
61. Rotate List
143. Reorder List
234. Palindrome Linked List
栈和队列的使用
20. Valid Parentheses
Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.
An input string is valid if:
- Open brackets must be closed by the same type of brackets.
- Open brackets must be closed in the correct order.
Note that an empty string is also considered valid.
public boolean isValid(String s) {
if(s == null || s.length() == 0) return true;
Stack<Character> stack = new Stack<Character>();
HashMap<Character, Character> map = new HashMap<Character, Character>();
map.put('(', ')'); map.put('[', ']'); map.put('{', '}');
for(int i = 0; i < s.length(); i++){
Character c = s.charAt(i);
if(c == '(' || c == '[' || c == '{') stack.push(c);
else if(stack.size() == 0) return false;
else {
Character c1 = map.get(stack.pop());
if (c != c1) return false;
}
}
return stack.empty();
}
150. Evaluate Reverse Polish Notation
71. Simplify Path
栈和递归
二叉树的递归：
前序遍历 144、
递归的方式
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> list = new ArrayList<Integer>();
if(root == null) return list;
recursivePreOrder(list, root);
return list;
}
private void recursivePreOrder(List<Integer> list, TreeNode root){
if(root == null) return;
list.add(root.val);
recursivePreOrder(list, root.left);
recursivePreOrder(list, root.right);
}
中序遍历 94、
递归的方式
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> list = new ArrayList<Integer>();
if(root == null) return list;
recursivePreOrder(list, root);
return list;
}
private void recursivePreOrder(List<Integer> list, TreeNode root){
if(root == null) return;
recursivePreOrder(list, root.left);
list.add(root.val);
recursivePreOrder(list, root.right);
}
后序遍历 145
递归的方式
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> list = new ArrayList<Integer>();
if(root == null) return list;
recursivePreOrder(list, root);
return list;
}
private void recursivePreOrder(List<Integer> list, TreeNode root){
if(root == null) return;
recursivePreOrder(list, root.left);
recursivePreOrder(list, root.right);
list.add(root.val);
}
341. Flatten Nested List Iterator
队列
广度优先遍历
102. Binary Tree Level Order Traversal
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> lists = new ArrayList<List<Integer>>();
if(root == null) return lists;
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.add(root);
while(!queue.isEmpty()){
List<Integer> list = new ArrayList<Integer>();
int levelSize = queue.size();
for(int i = 0; i < levelSize; i++){
TreeNode node = queue.poll();
if(node.left != null) queue.add(node.left);
if(node.right != null) queue.add(node.right);
list.add(node.val);
}
lists.add(list);
}
return lists;
}
107. Binary Tree Level Order Traversal II
103. Binary Tree Zigzag Level Order Traversal
199. Binary Tree Right Side View
BFS和图的最短路径
279. Perfect Squares
Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.
class Solution {
// 不是我写的
public int numSquares(int n) {
Queue<Node> queue = new LinkedList<Node>();
queue.add(new Node(n, 0));
int[] visit = new int[n+1];
while(!queue.isEmpty()){
Node node = queue.poll();
int num = node.num;
int step = node.step;
for(int i = 0; ; i++){
int a = num - i * i;
if(a < 0) break;
if(a == 0) return step + 1;
if(visit[a] == 1) continue;
queue.add(new Node(num - i * i, step + 1));
visit[a] = 1;
}
}
return -1;
}
}
class Node{
int num;
int step;
public Node(int num, int step){
this.num = num;
this.step = step;
}
}
127. Word Ladder
126. Word Ladder II
优先队列
347. Top K Frequent Elements
23. Merge k Sorted Lists