Sword refers to Offer 19. Regular expression matching (dynamic programming)

Thursday, February 11, 2021, the weather is fine [Do not lament the past, do not waste the present, do not fear the future]

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1. Introduction

Sword refers to Offer 19. Regular expression matching

2. Dynamic programming

This question is still very difficult. I just looked at the solution of this big guy directly. I feel that as long as I spend some time carefully thinking about it, the idea can still be clear. Here is the code:

class Solution {
    
    
public:
    bool isMatch(string s, string p) {
    
    
        int m = s.size();
        int n = p.size();
        vector<vector<bool>> dp(m+1,vector<bool>(n+1,false));
        for(int i=0;i<=m;++i){
    
    
            for(int j=0;j<=n;++j){
    
    
                // 分成空正则和非空正则两种
                if(j==0) dp[i][j] = i==0;
                // 非空正则分为两种情况 * 和 非 *
                // 非 * 的情况（当j=n时，j-1表示正则串p的最后一个字符）
                else if(p[j-1]!='*'){
    
    
                    if(i>0 && (p[j-1]==s[i-1] || p[j-1]=='.')){
    
    
                        dp[i][j] = dp[i-1][j-1];
                    }
                }
                // * 的情况
                else if(p[j-1]=='*'){
    
    
                    // 碰到 * 了，分为考虑和不考虑两种情况

                    // 不考虑 *
                    if(j>=2){
    
    
                        dp[i][j] = dp[i][j-2];
                    }
                    // 考虑 *（当i=m时，i-1表示字符串s的最后一个字符））
                    if(i>=1 && j>=2 && (s[i-1]==p[j-2] || p[j-2]=='.')){
    
    
                        dp[i][j] = dp[i][j] || dp[i-1][j];
                    }
                }    
            }
        }
        return dp[m][n];
    }
};

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references

"Sword Finger Offer Second Edition"

https://leetcode-cn.com/problems/zheng-ze-biao-da-shi-pi-pei-lcof/solution/zhu-xing-xiang-xi-jiang-jie-you-qian-ru-shen-by-je/