44. Wildcard Matching

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1刷
难题，第一次在leetcode刷了一道最像acm的题，知道用dp但是想不出来，s和p的遍历先后反了，导致公式写错了很多，哎！！！！

P[i][j] = P[i - 1][j - 1] && (s[i - 1] == p[j - 1] || p[j - 1] == ‘?’), if p[j - 1] != ‘*’;
P[i][j] = P[i][j - 1] || P[i - 1][j], if p[j - 1] == ‘*’.
2刷要重新刷一次dp，传说中还有贪心的算法，2刷的时候也要刷

Vector初始化vector

class Solution {public:    bool isMatch(string s, string p) {        vector<vector<bool>> dp(s.length() + 1, vector<bool>(p.length() + 1, false));        dp[0][0] = true;        for(int i = 1; i <= p.length(); ++ i){            if(p[i - 1] == '*') dp[0][i] = dp[0][i - 1];        }        for(int i = 1; i <= s.length(); ++ i){            for(int j = 1; j <= p.length(); ++ j){                if(p[j - 1] == '*'){                    if(dp[i - 1][j] || dp[i][j - 1]) dp[i][j] = true;                }else if(dp[i - 1][j - 1] && (s[i - 1] == p[j - 1] || p[j - 1] == '?')){                    dp[i][j] = true;                }            }        }        return dp[s.length()][p.length()];    }};

2刷
2刷的感觉好很多，能够想到基本思路和dp方程！！！！继续3刷这种题，肯定稳健，dp方程在1刷的题解

class Solution {public:    bool isMatch(string s, string p) {        int n = s.length();        int m = p.length();        vector<vector<bool>>dp(n + 1, vector<bool>(m + 1, false));        dp[0][0] = true;        for(int i = 1; i <= m; ++ i)            if(p[i - 1] == '*')                dp[0][i] = dp[0][i - 1];        for(int i = 1; i <= n; ++ i){            for(int j = 1; j <= m; ++ j){                if(p[j - 1] != '*'){                    if(s[i - 1] == p[j - 1] || p[j - 1] == '?')                        dp[i][j] = dp[i - 1][j - 1];                }                else{                    dp[i][j] = dp[i][j - 1] || dp[i - 1][j];                }            }        }        return dp[n][m];    }};

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