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; }