792. 匹配子序列的单词数

给定字符串 s 和字符串数组 words, 返回  words[i] 中是s的子序列的单词个数 。

字符串的 子序列 是从原始字符串中生成的新字符串，可以从中删去一些字符(可以是none)，而不改变其余字符的相对顺序。

例如， “ace” 是 “abcde” 的子序列。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/number-of-matching-subsequences
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

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双指针

class Solution {

    private boolean match(String str1, String str2) {
        int p1 = 0, p2 = 0;
        while (p1 < str1.length() && p2 < str2.length()) {
            if (str1.charAt(p1) == str2.charAt(p2)) {
                p1++;
                p2++;
            } else {
                p1++;
            }
        }
        return p2 == str2.length();
    }

    public int numMatchingSubseq(String s, String[] words) {
        int ans = 0;
        for (String word : words) {
            if (match(s, word)) {
                ans++;
            }
        }
        return ans;
    }
}

二分

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

class Solution {

    private int lower(List<Integer> data, int target) {
        int left = 0, right = data.size() - 1;
        while (left <= right) {
            int mid = (left + right) / 2;
            if (data.get(mid) < target) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        return left == data.size() ? -1 : left;
    }

    public int numMatchingSubseq(String s, String[] words) {
        if (s == null || words == null || words.length == 0) {
            return 0;
        }
        List<Integer>[] helper = new List[26];
        for (int i = 0; i < 26; ++i) {
            helper[i] = new ArrayList<>();
        }
        for (int i = 0; i < s.length(); ++i) {
            helper[s.charAt(i) - 'a'].add(i);
        }

        int ans = 0;

        for (String word : words) {
            int i;
            int target = 0;
            for (i = 0; i < word.length(); ++i) {
                List<Integer> arr = helper[word.charAt(i) - 'a'];
                int index = lower(arr, target);
                if (index == -1) {
                    break;
                }
                target = arr.get(index) + 1;
            }
            if (i == word.length()) {
                ans++;
            }
        }

        return ans;
    }
}

同时匹配

import java.util.ArrayDeque;
import java.util.Queue;

class Solution {
    public int numMatchingSubseq(String s, String[] words) {
        Queue<int[]>[] queue = new Queue[26];
        for (int i = 0; i < 26; ++i) {
            queue[i] = new ArrayDeque<>();
        }
        for (int i = 0; i < words.length; ++i) {
            queue[words[i].charAt(0) - 'a'].offer(new int[]{i, 0});
        }
        int ans = 0;
        for (int i = 0; i < s.length(); ++i) {
            int x = s.charAt(i) - 'a';
            int len = queue[x].size();
            while (len-- > 0) {
                int[] p = queue[x].poll();
                if (p[1] == words[p[0]].length() - 1) {
                    ans++;
                } else {
                    p[1]++;
                    queue[words[p[0]].charAt(p[1]) - 'a'].offer(p);
                }
            }
        }
        return ans;
    }
}

动态规划

import java.util.ArrayList;
import java.util.List;

class Solution {
    public int numMatchingSubseq(String t, String[] words) {
        int[][] dp = new int[t.length()][26];
        for (int i = t.length() - 1; i >= 0; i--) {
            for (int j = 0; j < 26; j++) {
                if (t.charAt(i) - 'a' == j) {
                    dp[i][j] = i;
                } else {
                    dp[i][j] = (i == t.length() - 1) ? -1 : dp[i + 1][j];
                }
            }
        }
        int ans = 0;

        for (String s: words) {
            int idx = 0, next = 0;
            while (idx < s.length() && next < t.length()) {
                next = dp[next][s.charAt(idx) - 'a'];
                if (next == -1) {
                    break;
                }
                idx ++;
                next ++;
            }
            ans += idx == s.length() ? 1 : 0;
        }

        
        return ans;
    }
}