题目描述：
Implement wildcard pattern matching with support for '?' and '*'.
'?' Matches any single character.
'*' Matches any sequence of characters (including the empty sequence).
The matching should cover the entire input string (not partial).
The function prototype should be:
bool isMatch(const char *s, const char *p)
Some examples:
isMatch("aa","a") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "*") → true
isMatch("aa", "a*") → true
isMatch("ab", "?*") → true
isMatch("aab", "c*a*b") → false
实现字符串匹配。输入字符串s,和字符串匹配p，?可被替换为任何非空字符，*可替换空字符或任何字符。
设，arr[i,j]表示s的第i个字符是否和p的第j个字符匹配。对于s[i]和p[j]，分为三种情况：
arr[i,j] = arr[i-1,j-1] ,p[j]=?
arr[i,j] = arr[i-1,j-1] && s[i-1] == p[j-1] ，p[j]为普通字符
arr[i,j] = arr[i-1,j-1]||arr[i-1,j]||arr[i,j-1] ，p[j]为*
实现代码：
Implement wildcard pattern matching with support for '?' and '*'.
'?' Matches any single character.
'*' Matches any sequence of characters (including the empty sequence).
The matching should cover the entire input string (not partial).
The function prototype should be:
bool isMatch(const char *s, const char *p)
Some examples:
isMatch("aa","a") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "*") → true
isMatch("aa", "a*") → true
isMatch("ab", "?*") → true
isMatch("aab", "c*a*b") → false
实现字符串匹配。输入字符串s,和字符串匹配p，?可被替换为任何非空字符，*可替换空字符或任何字符。
设，arr[i,j]表示s的第i个字符是否和p的第j个字符匹配。对于s[i]和p[j]，分为三种情况：
arr[i,j] = arr[i-1,j-1] ,p[j]=?
arr[i,j] = arr[i-1,j-1] && s[i-1] == p[j-1] ，p[j]为普通字符
arr[i,j] = arr[i-1,j-1]||arr[i-1,j]||arr[i,j-1] ，p[j]为*
实现代码：
public class Solution {
public bool IsMatch(string s, string p) {
var dp = new bool[s.Length + 1, p.Length + 1];
dp[0,0] = true; // s is empty , pattern is empty, match
// s is not empty , patter is empty , not match
for (var i = 0;i < s.Length; i++){
dp[i+1,0] = false;
}
// pattern not empty, s is empty , not match
for (var i = 0;i < p.Length; i++){
dp[0, i+1] = p[i] == '*' && dp[0, i];
}
for (var i = 1; i <= s.Length; i++){
for (var j = 1;j <= p.Length; j++){
if (p[j-1] == '?'){
dp[i,j] = dp[i-1,j-1]; // depends on previous match or no
}
else if(p[j-1] == '*'){
// 1. ab a*
// 2. bavfdc b*
// pattern j matches string i - 1 (* is any char)
// or
// pattern j-1 matches string i (* can be removed)
//Console.WriteLine(i+","+j);
dp[i,j] = dp[i-1,j] || dp[i, j-1] || dp[i-1,j-1];
}
else{
// pattern is a normal charactor , previous match also current char should match
dp[i,j] = dp[i-1,j-1] && s[i-1] == p[j-1];
}
}
}
//Console.WriteLine(dp);
return dp[s.Length,p.Length];
}
}
