标签：one this public sequence 大于 point 移动 boolean 相同
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 sand 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?
M1: two pointers
在t的长度范围内，比较两个指针指向的字母是否相同。如果相同，移动s指针，如果s到末尾了，返回true，无论是否相同，每次循环都移动t指针。如果t也走到末尾了，返回false
time: O(n), space: O(1)
class Solution { public boolean isSubsequence(String s, String t) { if(s.length() == 0) { return true; } int i = 0, j = 0; while(j < t.length()) { if(s.charAt(i) == t.charAt(j)) { i++; if(i == s.length()) { return true; } } j++; } return false; } }
M2: follow-up: binary search
需要对大量的s来check t，preprocess t 并存下信息比较合理。可以用hashmap存，遍历t并把每个字母作为key存入hashmap，value是字母出现的索引(list)。然后遍历s，对于s中的每个字母，先取出map中对应的下标list，再用binary search查找 是否存在大于当前字母前一个字母的下标(prev)的 当前字母的出现位置，如果不存在直接返回false，如果存在，prev自增1，直到遍历完s
time: O(MKlogN) -- M: average length of s, K: number of s, N: length of t (assuming all chars in t are the same)
space: O(length of t)
class Solution { public boolean isSubsequence(String s, String t) { Map<Character, List<Integer>> map = new HashMap<>(); for(int i = 0; i < t.length(); i++) { map.putIfAbsent(t.charAt(i), new ArrayList<>()); map.get(t.charAt(i)).add(i); } int prev = -1; for(int i = 0; i < s.length(); i++) { List<Integer> list = map.get(s.charAt(i)); if(list == null) { return false; } else { prev = binarySearch(list, prev); if(prev == -1) { return false; } prev++; } } return true; } public int binarySearch(List<Integer> list, int target) { int left = 0, right = list.size() - 1; while(left <= right) { int mid = left + (right - left) / 2; if(list.get(mid) < target) { left = mid + 1; } else { right = mid - 1; } } return left == list.size() ? -1 : list.get(left); } }
标签：one this public sequence 大于 point 移动 boolean 相同
原文地址：https://www.cnblogs.com/fatttcat/p/10245867.html
