leetcode 392. Is Subsequence

Given a string s and a string t, check if s is subsequence of t.You may assume that there is only lower case English letters in both s and t. t is potentially a very long (length ~= 500,000) string, and s is a short string (<=100).A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ace" is a subsequence of "abcde" while "aec" is not).Example 1:s = "abc", t = "ahbgdc"Return true.Example 2:s = "axc", t = "ahbgdc"Return false.Follow up:If there are lots of incoming S, say S1, S2, ... , Sk where k >= 1B, and you want to check one by one to see if T has its subsequence. In this scenario, how would you change your code?

最直观有效的解法当然是两个指针了

public class Solution {    public boolean isSubsequence(String s, String t) {        if(s.length()==0) return true;        int s_idx = 0, t_idx = 0;        while(s_idx < s.length() && t_idx < t.length()){            if(s.charAt(s_idx)==t.charAt(t_idx)){                s_idx++;                if(s_idx==s.length()) return true;            }            t_idx++;        }        return false;    }}

看到有人优化这个过程，用indexOf()函数

public class Solution {    public boolean isSubsequence(String s, String t)     {        if(t.length() < s.length()) return false;        int prev = 0;        for(int i = 0; i < s.length();i++)        {            char tempChar = s.charAt(i);            prev = t.indexOf(tempChar,prev);            if(prev == -1) return false;            prev++;        }        return true;    }}

其中indexOf的原理也不太懂。

对于follow up问题，这种问题一般都需要存一个map或之类的，存住一些中间结果。这题其实就是要存住在t串中每个字符出现的位置。类似一个散列表的结构。
然后遍历s的每个元素，查找该元素是否出现在t中，如果没有出现直接返回false,出现的话，要记住出现的位置，下次查找是，要大于该位置，即保持位置的顺序一致性。

    // Follow-up: O(N) time for pre-processing, O(Mlog?) for each S.    // Eg-1. s="abc", t="bahbgdca"    // idx=[a={1,7}, b={0,3}, c={6}]    //  i=0 ('a'): prev=1    //  i=1 ('b'): prev=3    //  i=2 ('c'): prev=6 (return true)    // Eg-2. s="abc", t="bahgdcb"    // idx=[a={1}, b={0,6}, c={5}]    //  i=0 ('a'): prev=1    //  i=1 ('b'): prev=6    //  i=2 ('c'): prev=? (return false)    public boolean isSubsequence(String s, String t) {        List<Integer>[] idx = new List[256]; // Just for clarity        for (int i = 0; i < t.length(); i++) {            if (idx[t.charAt(i)] == null)                idx[t.charAt(i)] = new ArrayList<>();            idx[t.charAt(i)].add(i);        }        int prev = 0;        for (int i = 0; i < s.length(); i++) {            if (idx[s.charAt(i)] == null) return false; // Note: char of S does NOT exist in T causing NPE            int j = Collections.binarySearch(idx[s.charAt(i)], prev);            if (j < 0) j = -j - 1;            if (j == idx[s.charAt(i)].size()) return false;            prev = idx[s.charAt(i)].get(j) + 1;        }        return true;    }