一、递归

题目描述

Given two binary trees, write a function to check if they are equal or not.
Two binary trees are considered equal if they are structurally identical and the nodes have the same value.

思路

• 碰到二叉树的问题，差不多就是深搜、广搜，递归那方面想想了，当然如果要考虑一下空间、时间，还需要进行剪枝和压缩处理。这题比较简单：判断两个树是否相等，可以递归的判断子树是否相等，最后找到边界条件就是是否都为空，都不为空时节点里面的值是否相等。

代码

``````/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isSameTree(TreeNode p, TreeNode q) {
if(p==null&&q==null){
return true;
}
else if(p==null||q==null){
return false;
}
else if(q.val!=p.val){
return false;
}
else{
return isSameTree(p.left,q.left)&&isSameTree(p.right,q.right);
}
}
}复制代码``````

二、动态规划

题目描述

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.
For example, given
s ="leetcode ",
dict =["leet ", "code "].
Return true because"leetcode "can be segmented as"leet code ".

思路

• dp的题目，写了几道了。核心无非就是确定dp数组和状态转移方程。这几道题都有明显的特点，那就是dp数组记录的就是所求的答案，所以答案一般都是dp[s.length()]这种形式的。
• 有了上面的总结，再来看着道题目。要求一串字母是否可以由所给的字典中的单词拼出来，要求返回布尔型。那好，也同时提示我们了dp数组就是记录它的子串是否能满足要求，类型是布尔型：dp[i]表示的是s[0,i-1]这个子串能否满足要求。
• dp[i]=dp[j]&&s[j,i]是否在字典中(0<=j<=i-1)

代码

``````import java.util.Set;

public class Solution {
public boolean wordBreak(String s, Set<String> dict) {
if(s==null||s.length()==0||dict==null||dict.size()==0){
return false;
}
boolean [] dp = new boolean [s.length()+2];
dp[0] = true;

for(int i=1;i<=s.length();i++){
for(int j=i-1;j>=0;j--){//从尾部扫描单词
if(dp[j]==true&&dict.contains(s.substring(j,i))){
dp[i]=true;
break;
}
else{
dp[i]=false;
}
}
}
return dp[s.length()];
}
}复制代码``````

题目描述

Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.
Return all such possible sentences.
For example, given
s ="catsanddog ",
dict =["cat ", "cats ", "and ", "sand ", "dog "].
A solution is["cats and dog ", "cat sand dog "].

思路

• 题目是上一道题的加强版，为了考察动态规划。感觉有点复杂，没想出来怎么在上一道的基础上改进。
• 深搜可以解决问题！直接看代码了

代码

``````import java.util.ArrayList;
import java.util.Set;

public class Solution {
public ArrayList<String> res = new ArrayList<>();
public ArrayList<String> wordBreak(String s, Set<String> dict) {
dfs(s,s.length(),dict," ");
return res;
}
public void dfs(String s,int index,Set<String> dict,String temp){
if(index==0){
if(temp.length()>0){