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+89
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code/lc100.java

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package code;
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/*
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* 100. Same Tree
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* 题意:判断两棵树是否一样
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* 难度:Easy
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* 分类:
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* 思路:
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* Tips:
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*/
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public class lc100 {
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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 isSameTree(TreeNode p, TreeNode q) {
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if(p==null&&q==null) return true;
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if( (p==null&&q!=null)||(q==null&&p!=null) ) return false;
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if(p.val!=q.val) return false;
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return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
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}
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}

code/lc119.java

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package code;
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import java.util.ArrayList;
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import java.util.List;
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/*
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* 119. Pascal's Triangle II
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* 题意:和118一样,就是输出不一样
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* 难度:Easy
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* 分类:Array
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* 思路:记一下 ArrayList.set 方法,不用重新 new ArrayList
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* Tips:
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*/
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public class lc119 {
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public List<Integer> getRow(int rowIndex) {
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ArrayList<Integer> res = new ArrayList<>();
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res.add(0,1);
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while(rowIndex-->0){
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res.add(0,1);
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for (int i = 1; i < res.size()-1 ; i++) {
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res.set(i, res.get(i)+res.get(i+1));
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}
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}
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return res;
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}
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}

code/lc14.java

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import java.util.Arrays;
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public class lc14 {
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public static void main(String[] args) {
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}
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public static String longestCommonPrefix(String[] strs) { //不是最优的方法,多做了比较
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if(strs.length==0)
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return "";

code/lc17.java

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* 17. Letter Combinations of a Phone Number
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* 题意:手机键盘字母输入
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* 难度:Medium
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* 思路:思路记一下,每次放到vector中,遍历vector,加上所有的可能
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* 分类:String, Backtracking
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*/
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public class lc17 {

code/lc19.java

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* 题意:删除链表中倒数第n个节点
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* 难度:Medium
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* 分类:Linked List, Two Pointers
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* 思路:快慢指针,快指针达到链表尾部时,满指针所在位置即为操作的节点
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* 思路:快慢指针,快指针达到链表尾部时,慢指针所在位置即为操作的节点
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* 注意:看清题意,是倒数第n个,且复杂度为n
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*/
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public class lc19 {

code/lc2.java

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* 注意考虑两个链表长度不一致的问题
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*/
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public class lc2 {
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public static void main(String[] args) {
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}
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public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
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int sum = 0;

code/lc26.java

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* 题意:移除数组中重复元素
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* 难度:Easy
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* 分类:Array, Two Pointers
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* lc27
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*/
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public class lc26 {
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public static void main(String[] args) {

code/lc27.java

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* 分类:Array, Two Pointers
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* 思路:两个指针,分别O(n),指向要交换的位置和和他交换的数
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* 答案中直接遍历一遍数组,放到位置上就行了,i++
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* Tips:
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* Tips:lc26
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*/
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public class lc27 {
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public int removeElement(int[] nums, int val) {

code/lc28.java

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* 难度:Easy
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* 分类:Two Pointers, String
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* Tips:注意判断子串为空的方法为needle.length()==0,不要用needle==""
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* 最优的解法应该是O(N)的,类似KMP的思路,不过面试不会让写KMP的
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*/
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public class lc28 {
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public static void main(String[] args) {

code/lc29.java

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* 难度:Medium
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* 分类:Math, Binary Search
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* 思路:被除数减去除数,除数每次左移一位,也就是*2 来实现类似二分的思想
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* Tips:注意下用long类型,以及溢出的情况
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* Tips:注意下用long类型,以及溢出的情况,注意符号
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*/
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public class lc29 {
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public static void main(String[] args) {

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