LintCode 1263 · Is Subsequence (双指针好题)

1263 · Is Subsequence
Algorithms
Description
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 (length <= 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
Example 1:

Input: s = “abc”, t = “ahbgdc”
Output: true
Example 2:

Input: s = "axc", t = "ahbgdc"
Output: false
Challenge
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?

解法1: 双指针。时间复杂度O(n)。空间复杂度O(1)。

class Solution {
public:
    /**
     * @param s: the given string s
     * @param t: the given string t
     * @return: check if s is subsequence of t
     */
    bool isSubsequence(string &s, string &t) {
        if (s.size() > t.size()) return false;
        int indexS = 0, indexT = 0;
        while (indexT < t.size()) {
            if (s[indexS] == t[indexT]) {
                indexS++;
            }
            indexT++;
        }
        return indexS == s.size();
    }
};

解法2：DP。时间复杂度O(n^2)，空间复杂度O(n)。

class Solution {
public:
    /**
     * @param s: the given string s
     * @param t: the given string t
     * @return: check if s is subsequence of t
     */
    bool isSubsequence(string &s, string &t) {
        int sizeS = s.size(), sizeT = t.size();
        if (sizeS > sizeT) return false;
       
        vector<vector<int>> dp(2, vector<int>(sizeT + 1));
        for (int i = 1; i <= sizeS; i++) {
            for (int j = 1; j <= sizeT; j++) {
                if (s[i - 1] == t[j - 1]) {
                    dp[i % 2][j] = dp[(i - 1) % 2][j - 1] + 1;
                } else {
                    dp[i % 2][j] = max(dp[(i - 1) % 2][j], dp[i % 2][j - 1]);
                }
            }
        }
        return dp[sizeS % 2][sizeT] == sizeS;
    }
};