leetcode新年病房暴乱康复计划 10.正则表达式匹配 DP

/**
 * @param {string} s
 * @param {string} p
 * @return {boolean}
 */
var isMatch = function(s, p) {
    let dp = [];
    let len = s.length;
    let _len = p.length;
    for(let i = 0; i <= len; i++){
        dp.push([]);
        dp[i][0] = 0;
    }
    dp[0][0] = 1;
    for(let i = 0; i <= _len; i++){
        if(p.charAt(i) === ‘*‘ && dp[0][i - 1]){
            dp[0][i + 1] = 1;
        }else{
            dp[0][i + 1] = 0;
        }
    }
    for(let i = 0; i < len; i++){
        for(let j = 0; j < _len; j++){
            if((s.charAt(i) === p.charAt(j) || p.charAt(j) === ‘.‘) && dp[i][j]){
                dp[i + 1][j + 1] = 1;
            }else if(p.charAt(j) === ‘*‘ && (dp[i + 1][j] || dp[i][j + 1]) && (s.charAt(i) === p.charAt(j - 1) || p.charAt(j - 1) === ‘.‘)){
                dp[i + 1][j + 1] = 1;
            }else if(p.charAt(j) === ‘*‘ && dp[i + 1][j - 1]){
                dp[i + 1][j + 1] = 1;
            }else{
                dp[i + 1][j + 1] = 0;
            }
        }
    }
    return dp[len][_len];
};

如果 p.charAt(j) == s.charAt(i) : dp[i][j] = dp[i-1][j-1]；
如果 p.charAt(j) == ‘.‘ : dp[i][j] = dp[i-1][j-1]；
如果 p.charAt(j) == ‘*‘：
　　如果 p.charAt(j-1) != s.charAt(i) : dp[i][j] = dp[i][j-2] //in this case, a* only counts as empty
　　如果 p.charAt(i-1) == s.charAt(i) or p.charAt(i-1) == ‘.‘：
　　　　dp[i][j] = dp[i-1][j] //in this case, a* counts as multiple a
　　　　or dp[i][j] = dp[i][j-1] // in this case, a* counts as single a
　　　　or dp[i][j] = dp[i][j-2] // in this case, a* counts as empty

来自乔碧萝殿下?的题解

原文地址：https://www.cnblogs.com/qq965921539/p/12283924.html