LeetCode Distinct Subsequences

基于Unidbg的小红书加密协议动态分析与算法还原实战 在移动应用安全领域,加密协议与签名机制是保障数据传输安全、实现客户端身份验证的核心技术。其原理通常涉及哈希算法、对称/非对称加密以及自定义的混淆逻辑,旨在防止请求被篡改或重放。理解这些协议对于安全研究、风控策略制定及合规的数据交互分析具有重要价值。在实际的工程实践中,面对高度混淆和加固的Native代码(SO库),传统的静态分析往往难以深入。此时,沙盒模拟执行技术提供了一种动态、可控的分析方案。通过搭建一个虚拟的移动应用运行环境,研究人员可以在无需真机、规避反调试机制的情况下,完整地追踪和观察加密算法的执行 阅读详情

原题链接在这里:https://leetcode.com/problems/distinct-subsequences/

这是一道DP. 简历二维数组dp, dp[i,j]表示s[0....i]有多少种方法可以减成t[0.......j].

更新方法是: 若s.charAt(i-1) == t.charAt(j-1), dp[i][j] = dp[i-1][j-1](用上最新的match, 之前s[0....i-1]减到t[0...j-1]的方法都用最后一个match, 所以总共的方法数不变) 加上dp[i-1][j](不用当前match, 之前s[0...i-1]减到t[0...j]的方法即可).

若s.charAt(i-1) != t.charAt(j-1),  不match的话只能是之前s[0...i-1]减到t[0...j]的方法。

初始化:若t的长度为0, s减到t的方法只有一种, 若s的长度为0,t的长度不为0,s减到t方法数为0.

所以dp第一行为1,第一列除了第一行为0.

Time O(lens*lent). Space O(lens*lent).

Note: string.charAt(index), index是从0开始的,所以要用s.charAt(i-1)比较,-1不可省略。

AC Java:

public class Solution {
    public int numDistinct(String s, String t) {
        if(t == null || t.length() == 0){
            return 1;
        }
        if((s == null || s.length() == 0) && (t != null && t.length() != 0)){
            return 0;
        }
        int lens = s.length();
        int lent = t.length();
        int[][] dp = new int[lens+1][lent+1];
        for(int i = 0; i<=lens; i++){
            dp[i][0] = 1;
        }
        for(int j = 1; j<=lent; j++){
            dp[0][j] = 0;
        }
        for(int i = 1; i<=lens; i++){
            for(int j = 1; j<=lent; j++){
                if(s.charAt(i-1) == t.charAt(j-1)){
                    dp[i][j] = dp[i-1][j-1] + dp[i-1][j];
                }else{
                    dp[i][j] = dp[i-1][j];
                }
            }
        }
        return dp[lens][lent];  //error
    }
}


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