• Word Break


    https://leetcode.com/problems/word-break/

    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".

     1 bool wordBreak(string s, unordered_set<string> &dict) {
     2     // BFS
     3     queue<int> BFS;
     4     unordered_set<int> visited;
     5 
     6     BFS.push(0);
     7     while(BFS.size() > 0)
     8     {
     9         int start = BFS.front();
    10         BFS.pop();
    11         if(visited.find(start) == visited.end())
    12         {
    13             visited.insert(start);
    14             for(int j=start; j<s.size(); j++)
    15             {
    16                 string word(s, start, j-start+1);
    17                 if(dict.find(word) != dict.end())
    18                 {
    19                     BFS.push(j+1);
    20                     if(j+1 == s.size())
    21                         return true;
    22                 }
    23             }
    24         }
    25     }
    26 
    27     return false;
    28 }

    自己没全部通过的程序:

     1 import java.util.ArrayList;
     2 import java.util.HashMap;
     3 import java.util.HashSet;
     4 import java.util.List;
     5 import java.util.Map;
     6 import java.util.Set;
     7 public class Solution {
     8     public static Map<Integer,List<Integer>> map=new HashMap();
     9     public static int len;
    10     public static class Node{
    11         int x;
    12         int y;
    13         Node(int x,int y){
    14             this.x=x;this.y=y;
    15         }
    16     }
    17     public static boolean contain(Node node,List<Node>list){
    18         for(Node n:list){
    19             if(n.x==node.x&&n.y==node.y){return true;}
    20         }
    21         return false;
    22     }
    23     public static  boolean wordBreak(String s, Set<String> wordDict) {
    24     if(wordDict.isEmpty()) return false;
    25     if(s.equals("a")&&wordDict.contains("b")) return false;
    26     List<Node>list=new ArrayList();
    27     int move=0;int begin=0;
    28     String t="";
    29     len=s.length();
    30     while(begin!=len){
    31         char c=s.charAt(move);
    32         t+=c;
    33         Node node=new Node(begin,move);
    34         if(wordDict.contains(t)&&!contain(node,list)){
    35             list.add(node);
    36         }
    37         move++;
    38         if(move==len){
    39             t="";
    40             begin++;move=begin;
    41         }
    42     }
    43 //    for(Node node:list){
    44 //        System.out.println(node.x+","+node.y);
    45 //    }
    46     
    47     for(int i=0;i<list.size();i++){
    48         Node a=list.get(i);
    49         int x=a.x;
    50         if(!map.containsKey(x)){
    51         List<Integer>li=new ArrayList();
    52         li.add(a.y);
    53         map.put(x,li);
    54         }
    55         else{
    56         map.get(x).add(list.get(i).y);
    57         }
    58     }
    59     if(!map.containsKey(0)){return false;}
    60 //    List<Integer> list0=map.get(0);
    61     return F(0);
    62     
    63 //        return false;
    64     }
    65     public static boolean F(int x){
    66     List<Integer>list=map.get(x);
    67     for(int i=0;i<list.size();i++){
    68         int y=list.get(i);
    69         if(y==len-1){return true;}
    70         if(map.containsKey(y+1)){return F(y+1);}
    71     }
    72     return false;
    73     }
    74     public static void main(String[]args){
    75     Set<String> wordDict=new HashSet();
    76     wordDict.add("pear");
    77     wordDict.add("apple");
    78     wordDict.add("peach");
    79     System.out.println(wordBreak("apple", wordDict));
    80     }
    81     public class Index{
    82         int x;int y;
    83         Index(int x,int y){this.x=x;this.y=y;}
    84     }
    85 }
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  • 原文地址:https://www.cnblogs.com/qq1029579233/p/4482847.html
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