LeetCode1268搜索推荐系统

题目描述

  给你一个产品数组 products 和一个字符串 searchWord ，products 数组中每个产品都是一个字符串。请你设计一个推荐系统，在依次输入单词 searchWord 的每一个字母后，推荐 products 数组中前缀与 searchWord 相同的最多三个产品。如果前缀相同的可推荐产品超过三个，请按字典序返回最小的三个。请你以二维列表的形式，返回在输入 searchWord 每个字母后相应的推荐产品的列表。

解析

  在上一题实现前缀树的基础上，只需要查询前缀树中有前缀的前三个单词即可，使用递归去查找子节点。

static class Trie {
        boolean isEndOfWord;
        Trie[] children;

        public Trie() {
            this.isEndOfWord = false;
            this.children = new Trie[26];
        }

        public Trie(String[] stringList) {
            this();
            for (String s : stringList) {
                this.insert(s);
            }
        }

        public void insert(String word) {
            Trie cur = this;
            for (int i = 0; i < word.length(); i++) {
                int index = word.charAt(i) - 'a';
                if (cur.children[index] == null) {
                    cur.children[index] = new Trie();
                }
                cur = cur.children[index];
            }
            cur.isEndOfWord = true;
        }

        private void DFS(List<String> res, Trie node, int num, StringBuilder str) {
            if (res.size() >= num) {
                return;
            }
            if (node.isEndOfWord) {
                res.add(str.toString());
            }
            for (int i = 0; i < 26; i++) {
                if (node.children[i] != null) {
                    str.append((char) (i + 'a'));
                    DFS(res, node.children[i], num, str);
                    str.deleteCharAt(str.length() - 1);
                }
            }
        }

        public List<String> getRecommendation(String prefix, int num) {
            List<String> res = new ArrayList<>(num);
            Trie cur = this;
            for (int i = 0; i < prefix.length(); i++) {
                int index = prefix.charAt(i) - 'a';
                if (cur.children[index] == null) {
                    return res;
                }
                cur = cur.children[index];
            }
            DFS(res, cur, num, new StringBuilder(prefix));
            return res;
        }
    }

    public List<List<String>> suggestedProducts(String[] products, String searchWord) {
        List<List<String>> res = new ArrayList<>(searchWord.length());
        Trie trie = new Trie(products);
        StringBuilder prefix = new StringBuilder();
        for (char c : searchWord.toCharArray()) {
            prefix.append(c);
            res.add(trie.getRecommendation(prefix.toString(), 3));
        }
        return res;
    }

  实际上，并不需要使用前缀树这种复杂的结构，我们可以直接将产品字符串数组按照ASCII码的顺序排序，然后通过二分查找去查询即可。查询使用String的compareTo方法即可，如果products[mid].compareTo(prefix) < 0，说明待搜索的元素在中值的右边，继续搜索右边即可。

public List<List<String>> suggestedProducts(String[] products, String searchWord) {
        Arrays.sort(products);
        List<List<String>> res = new ArrayList<>();
        String prefix = "";
        for (char c : searchWord.toCharArray()) {
            prefix += c;
            int start = binarySearch(products, prefix);
            List<String> recommendations = new ArrayList<>();
            for (int i = start; i < products.length && recommendations.size() < 3; i++) {
                if (products[i].startsWith(prefix)) {
                    recommendations.add(products[i]);
                } else {
                    break;
                }
            }
            res.add(recommendations);
        }
        return res;
    }

    private int binarySearch(String[] products, String prefix) {
        int low = 0, high = products.length;
        while (low < high) {
            int mid = (low + high) / 2;
            if (products[mid].compareTo(prefix) < 0) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }
        return low;
    }

  上面的方法使用了startsWith这个方法，对于测试用例来说，比较的字符串非常长，那么这种方式是非常耗时的，那么可以用两次二分查找去找前缀的起点和终点，得到区间就不需要使用startsWith去判断了。

private int findBorder(String[] a, int start, int end, int i, char s, boolean left) {
        while (start <= end) {
            int mid = (start + end) / 2;            
            if (a[mid].length() <= i) {
                start = mid + 1;
                continue;
            }            
            char c = a[mid].charAt(i);            
            if (c < s) {
                a[mid] = "";
                start = mid + 1;
            } else if (c > s) {
                end = mid - 1;
            } else if (left) {
                end = mid - 1;
            } else {
                start = mid + 1;
            }
        }
        
        if (left) {
            return start;
        } else {
            return end;
        }
    }
    
    private int findLeft(String[] a, int start, int end, int i, char s) {
        return findBorder(a, start, end, i, s, true);
    }
    
    private int findRight(String[] a, int start, int end, int i, char s) {
        return findBorder(a, start, end, i, s, false);
    }
    
    public List<List<String>> suggestedProducts(String[] products, String searchWord) {
        char[] s = searchWord.toCharArray();  
        Arrays.sort(products);      
        int n = products.length;
        int start = 0;
        int end = n - 1;        
        List<List<String>> output = new ArrayList<>(s.length);        
        for (int i = 0; i < s.length; i++) {
            output.add(new ArrayList<>());
        }        
        for (int i = 0; i < s.length; i++) {
            char c = s[i];
            int tmpStart = start;
            int tmpEnd = end;
            start = findLeft(products, tmpStart, tmpEnd, i, c);
            end = findRight(products, tmpStart, tmpEnd, i, c);            
            if (start > end) {
                break;
            }            
            for (int j = 0; j < Math.min(end - start + 1, 3); j++) {
                output.get(i).add(products[start + j]);
            }
        }        
        return output;
    }