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);
}
}
}
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()];
}
}
三、深搜
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){
res.add(temp.substring(0,temp.length()-1));//如果写成res.add(temp)则末尾会多一个空格,小细节
}
}
for(int i=index-1;i>=0;i--){
if(dict.contains(s.substring(i,index))){
dfs(s,i,dict,s.substring(i,index)+" "+temp);
}
}
}
}