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);
}
}
原文地址：https://www.cnblogs.com/fatttcat/p/10245867.html