题目

题目链接：
https://www.nowcoder.com/practice/c0d32c1ce5744472a01b2351a2c2767f

思路

前缀树+动态规划

参考答案Java

import java.util.*;


public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param s string字符串
     * @param dic string字符串一维数组
     * @return bool布尔型
     */
    public boolean wordDiv (String s, String[] dic) {
        //前缀树+动态规划
        PreTreeNode trie = new PreTreeNode();
        for (String s1 : dic) {
            add(trie, s1);
        }

        int n = s.length();
        boolean[] dp = new boolean[n + 1];
        dp[n] = true;
        for (int i = n - 1; i >= 0 ; i--) {
            PreTreeNode cur = trie;
            for (int j = i; j < n ; j++) {
                char c = s.charAt(j);
                cur = cur.nexts.get(c);
                if (cur == null) break;
                if (cur.end > 0) {
                    dp[i] = dp[i] | dp[j + 1];
                }
            }
        }
        return dp[0];
    }

    static class PreTreeNode {
        int pass = 0;
        int end = 0;
        Map<Character, PreTreeNode> nexts = new HashMap<>();
    }

    public void add(PreTreeNode root, String word) {
        if (word == null || word.length() == 0) return;
        PreTreeNode cur = root;
        cur.pass++;
        for (int i = 0; i < word.length() ; i++) {
            char c = word.charAt(i);
            if (!cur.nexts.containsKey(c))
                cur.nexts.put(c, new PreTreeNode());

            cur = cur.nexts.get(c);
            cur.pass++;
        }

        cur.end++;
    }
}

参考答案Go

package main

/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 *
 * @param s string字符串
 * @param dic string字符串一维数组
 * @return bool布尔型
 */
func wordDiv(s string, dic []string) bool {
	//前缀树+动态规划
	trie := &PreTreeNode{0, 0, map[byte]*PreTreeNode{}}
	for i := 0; i < len(dic); i++ {
		trie.Add(dic[i])
	}

	n := len(s)
	dp := make([]bool, n+1)
	dp[n] = true
	for i := n - 1; i >= 0; i-- {
		cur := trie
		for j := i; j < n; j++ {
			c := s[j]
			tmp, ok := cur.nexts[c]
			if !ok {
				break
			}
			cur = tmp

			if tmp.End > 0 {
				dp[i] = dp[i] || dp[j+1]
			}
		}
	}

	return dp[0]
}

type PreTreeNode struct {
	Pass int
	End  int

	nexts map[byte]*PreTreeNode
}

func (node *PreTreeNode) Add(word string) {
	if len(word) == 0 {
		return
	}

	cur := node
	cur.Pass++

	for i := 0; i < len(word); i++ {
		c := word[i]

		_, ok := cur.nexts[c]
		if !ok {
			cur.nexts[c] = &PreTreeNode{0, 0, map[byte]*PreTreeNode{}}
		}

		cur = cur.nexts[c]
		cur.Pass++
	}

	cur.End++

}

参考答案PHP

<?php


/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param s string字符串 
 * @param dic string字符串一维数组 
 * @return bool布尔型
 */
function wordDiv( $s ,  $dic )
{
    
   // 前缀树+动态规划
    $trie = new Node();
    foreach ($dic as $word){
        add($trie,$word);
    }

    $n = strlen($s);
    $dp= array();
    $dp[$n] = true;

    for($i=$n-1;$i>=0;$i--){
        $cur = $trie;
        for($j=$i;$j<$n;$j++){
            $c = $s[$j];
            $cur =$cur->nexts[$c];
            if($cur==null ||empty($cur))
                break;

            if($cur->end >0){
                $dp[$i]=$dp[$i] ||$dp[$j+1];
            }
        }
    }
    return $dp[0];
}

class Node{
    public $pass = 0;
    public $end =0;
    public $nexts = array();
}

//添加到单词到前缀树
function add(&$root,$word){
    $cur =$root;
    $cur->pass++;
    for($i=0;$i<strlen($word);$i++){
        $c = $word[$i];

        if(!isset($cur->nexts[$c])){
            $cur->nexts[$c] = new Node();
        }
        $cur = $cur->nexts[$c];
        $cur->pass++;
    }

    $cur->end++;
}