题目一
请实现一个函数用来匹配包括‘.‘和‘*‘的正则表达式。模式中的字符‘.‘表示任意一个字符，而‘*‘表示它前面的字符可以出现任意次（包含0次）。 在本题中，匹配是指字符串的所有字符匹配整个模式。例如，字符串"aaa"与模式"a.a"和"ab*ac*a"匹配，但是与"aa.a"和"ab*a"均不匹配
思路
用动态规划来解决
规定
若
，则str[0~i-1]和pattern[0~j-1]匹配
1、当p[j-1] != * 时
2、当p[j-1] = * 时
记p[j-2]=X,分为以下两种情况
(a) "X*"重复了0次：等式右边第一项说明*重复零次，则f[i][j]与 f[i][j-2]真假相同，亦或是p[j-2]是万能字符
(b)"X*"重复了大于等于1次：
等式右边第一项说明要么是万能字符，
第二项表示f[i][j]与f[i-1][j]真假相同，其实就是回退到重复0次的情况
第三项保证复制的正确性
如 s=bcaaa p=bca* f[5][4] -> f[4][4] -> f[3][4] -> f[2][4]
此时 s = bc p = bca*，可以匹配
算法流程
1、初始化二维布尔数组，注意列不一定为0
2、按以上规则自下网上计算出f[i][j]
计算过程
| 空 | b | c | a | a | a | |
| 空 | 1 | 0 | 0 | 0 | 0 | 0 |
| b | 0 | 1 | 0 | 0 | 0 | 0 |
| c | 0 | 0 | 1 | 0 | 0 | 0 |
| a | 0 | 0 | 0 | 1 | 0 | 0 |
| * | 0 | 0 | 1 | 1 | 1 | 1 |
class Solution {
public:
bool match(char* str, char* pattern)
{
int m = strlen(str);
int n = strlen(pattern);
vector<vector<bool>> f(m+1, vector<bool>(n+1, false));
f[0][0] = true;
for (int i = 1; i <= m; i++)
{
f[i][0] = false;
}
for (int j = 1; j <= n; j++)
{
f[0][j] = ((j > 1) && (pattern[j-1] == ‘*‘) && (f[0][j-2]));
}
for (int i = 1; i <= m; i++){
for (int j = 1; j <= n; j++){
if (pattern[j-1] != ‘*‘)
f[i][j] = f[i - 1][j - 1] && (str[i - 1] == pattern[j - 1] || ‘.‘ == pattern[j - 1]);
else
f[i][j] = f[i][j-2] || (pattern[j-2] == ‘.‘) || (str[i-1] == pattern[j-2]) && f[i-1][j];
}
}
return f[m][n];
}
};
题目二
Given an input string (s) and a pattern (p), implement wildcard pattern matching with support for ‘?‘ and ‘*‘.
‘?‘ Matches any single character. ‘*‘ Matches any sequence of characters (including the empty sequence).
Example 1:
Input: s = "aa" p = "a" Output: false Explanation: "a" does not match the entire string "aa".
Example 2:
Input: s = "aa" p = "*" Output: true Explanation: ‘*‘ matches any sequence.
Example 3:
Input: s = "cb" p = "?a" Output: false Explanation: ‘?‘ matches ‘c‘, but the second letter is ‘a‘, which does not match ‘b‘.
Example 4:
Input: s = "adceb" p = "*a*b" Output: true Explanation: The first ‘*‘ matches the empty sequence, while the second ‘*‘ matches the substring "dce".
Example 5:
Input: s = "acdcb" p = "a*c?b" Output: false
class Solution {
public:
bool isMatch(string s, string p) {
int n = (int)s.length();
int m = (int)p.length();
vector<vector<bool>> dp(n + 1, vector<bool>(m + 1));
dp[0][0] = true;
for (int i = 1; i <= m && p[i - 1] == ‘*‘; i++)
dp[0][i] = true;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (p[j - 1] == ‘*‘)
dp[i][j] = dp[i - 1][j] || dp[i][j - 1];
else
dp[i][j] = (p[j - 1] == ‘?‘ || s[i - 1] == p[j - 1]) && dp[i - 1][j - 1];
}
}
return dp[n][m];
}
};
原文地址：https://www.cnblogs.com/shiganquan/p/9340833.html