1. If p.charAt(j) == s.charAt(i) : dp[i][j] = dp[i-1][j-1];

2. If p.charAt(j) == ‘.’ : dp[i][j] = dp[i-1][j-1];

3. If p.charAt(j) == ‘*’:

   here are two sub conditions:

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

   2. if 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
   

const isMatch = function (s, p) {
  const m = s.length, n = p.length;
  // if (m === 0 || n === 0) {
  //   return false;
  // }
  let dp = new Array();
  for (let i = 0; i <= m; i++) {
    dp[i] = new Array();
    for (let j = 0; j <= n; j++) {
      dp[i][j] = false;
    }
  }
  dp[0][0] = true;
  for (let i = 0; i < n; i++) {
    if (p[i] === '*' && dp[0][i - 1]) {
      dp[0][i + 1] = true;
    }
  }
  for (let i = 0; i < m; i++) {
    for (let j = 0; j < n; j++) {
      if (p[j] === '.' || p[j] === s[i]) {
        dp[i + 1][j + 1] = dp[i][j];
      }
      if (p[j] === '*') {
        if (p[j - 1] !== s[i] && p[j - 1] !== '.') {
          dp[i + 1][j + 1] = dp[i + 1][j - 1];
        } else {
          // 依次为 'a*' === '' 'a*' === 'a' 'a*' === 'n * a'
          // 其实在当前条件分支中已经判定了 'a*' === 'a' 的情况了 ,所以下面的第二个判断可以删掉
          dp[i + 1][j + 1] = dp[i + 1][j - 1] || dp[i + 1][j] || dp[i][j + 1];
        }
      }
    }
  }
  return dp[m][n];
};