Is Subsequence
Given two strings s and t, return true if s is a subsequence of t, or false otherwise.
A subsequence of a string is a new string that is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (i.e., "ace" is a subsequence of "abcde" while "aec" is not).
Example 1:
Input: s = "abc", t = "ahbgdc"
Output: true
Example 2:
Input: s = "axc", t = "ahbgdc"
Output: false
Constraints:
0 <= s.length <= 100
0 <= t.length <= 104
s and t consist only of lowercase English letters.
Follow up: Suppose there are lots of incoming s, say s1, s2, ..., sk where k >= 109, and you want to check one by one to see if t has its subsequence. In this scenario, how would you change your code?
思路一：双层循环遍历，注意遍历 t 的每个字符时，起始位置需要额外记录
public boolean isSubsequence(String s, String t) {
int idx = 0;
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
boolean find = false;
while (idx < t.length()) {
if (t.charAt(idx++) == c) {
find = true;
break;
}
}
if (!find) return false;
}
return true;
}
思路二：双指针，思路巧妙，通过两个指针递增的方式实现遍历对比
public boolean isSubsequence(String s, String t) {
int n = s.length(), m = t.length();
int i = 0, j = 0;
while (i < n && j < m) {
if (s.charAt(i) == t.charAt(j)) {
i++;
}
j++;
}
return i == n;
}
思路三：动态规划，提前处理字符串，记录字符出现的位置，利用空间换时间
public boolean isSubsequence(String s, String t) {
int n = s.length(), m = t.length();
int[][] f = new int[m + 1][26];
for (int i = 0; i < 26; i++) {
f[m][i] = m;
}
for (int i = m - 1; i >= 0; i--) {
for (int j = 0; j < 26; j++) {
if (t.charAt(i) == j + 'a') f[i][j] = i;
else f[i][j] = f[i + 1][j];
}
}
int add = 0;
for (int i = 0; i < n; i++) {
if (f[add][s.charAt(i) - 'a'] == m) {
return false;
}
add = f[add][s.charAt(i) - 'a'] + 1;
}
return true;
}