关于“经常调用怎么办”这类的follow up

Reverse bits of a given 32 bits unsigned integer.
Example:
Input: 43261596
Output: 964176192
Explanation: 43261596 represented in binary as 00000010100101000001111010011100,
return 964176192 represented in binary as 00111001011110000010100101000000.
Follow up:
If this function is called many times, how would you optimize it?

// cache
private final Map cache = new HashMap();
public int reverseBits(int n) {
    byte[] bytes = new byte[4];
    for (int i = 0; i < 4; i++) // convert int into 4 bytes
        bytes[i] = (byte)((n >>> 8*i) & 0xFF);
    int result = 0;
    for (int i = 0; i < 4; i++) {
        result += reverseByte(bytes[i]); // reverse per byte
        if (i < 3)
            result <<= 8;
    }
    return result;
}

private int reverseByte(byte b) {
    Integer value = cache.get(b); // first look up from cache
    if (value != null)
        return value;
    value = 0;
    // reverse by bit
    for (int i = 0; i < 8; i++) {
        value += ((b >>> i) & 1);
        if (i < 7)
            value <<= 1;
    }
    cache.put(b, value);
    return value;
}

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?

public boolean isSubsequence(String s, String t) {
        List[] 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;
    }

注意：对于这类问题，首先可以想到cache，有些东西算一遍之后可以重复使用。