问题分析

这道题的难点主要在于*号的匹配，这里记dp[i][j]表示s[1...i]和p[1...j]能否完成匹配，先根据特殊情况归纳总结：

• *号匹配 0 次，则dp[i][j] = dp[i][j-2]
• *号匹配 1 次，则dp[i][j] = dp[i-1][j-2] && s[i] == p[j-1]
• *号匹配 2 次，则dp[i][j] = dp[i-2][j-2] && s[i-1] == p[j-1] && s[i] == p[j-1]
• *号匹配 3 次，则dp[i][j] = dp[i-3][j-2] && ...
• *号匹配 k 次，则dp[i][j] = dp[i-k][j-2] && ...

对上述进行归纳可得：

• 当k = 0时，dp[i][j] = dp[i][j-2]
• 当k >= 1时，将dp[i][j] = dp[i][j-2]带入等式左边，可得dp[i][j-2] = dp[i-k][j-2] && ...。因此，最终化简可得dp[i][j] = dp[i-1][j] && s[i] == p[j-1]

算法描述

状态定义：dp[i][j]表示s[1...i]和p[1...j]能否完成匹配

状态转移：
若p[j] == '*'，可能的状态转移如下：

• *表示匹配 0 个前面的那一个元素：dp[i][j] = dp[i][j-2]
• *表示匹配多个前面的那一个元素：dp[i][j] = (s[i] == p[j-1] || p[j-1] == '.') && dp[i-1][j]
  若p[j] != '*'，则只能进行单元素的匹配：dp[i][j] = (s[i] == p[j] || p[j] == '.') && dp[i-1][j-1]

边界情况：

• dp[0][0] = true
• 子字符规律p[1...i]可能匹配空字符串s：dp[0][i] = p[i] == '*' && dp[0][i-2]

程序代码

C++

class Solution {
public:bool isMatch(string s, string p) {int m = s.size(), n = p.size();s = " " + s;p = " " + p;vector<vector<bool>> dp(m + 1, vector<bool>(n + 1, false));// 初始化dp[0][0] = true;for(int i = 2; i <= n; i++)  dp[0][i] = p[i] == '*' && dp[0][i-2];for(int i = 1; i <= m; i++) {for(int j = 1; j <= n; j++) {if(p[j] == '*') {dp[i][j] = dp[i][j-2] || (s[i] == p[j-1] || p[j-1] == '.') && dp[i-1][j];}else {dp[i][j] = (s[i] == p[j] || p[j] == '.') && dp[i-1][j-1];}}}return dp[m][n];}
};

Go

func isMatch(s string, p string) bool {m, n := len(s), len(p)s = " " + sp = " " + pdp := make([][]bool, m + 1)for i := 0; i < len(dp); i++ {dp[i] = make([]bool, n + 1)}// 初始化dp[0][0] = truefor i := 2; i <= n; i++ {dp[0][i] = dp[0][i-2] && p[i] == '*'}for i := 1; i <= m; i++ {for j := 1; j <= n; j++ {if p[j] == '*' {dp[i][j] = dp[i][j-2] || dp[i-1][j] && (s[i] == p[j-1] || p[j-1] == '.')} else {dp[i][j] = dp[i-1][j-1] && (s[i] == p[j] || p[j] == '.')}}}return dp[m][n]
}