Given an input string (s) and a pattern (p), implement wildcard pattern matching with support for '?' and '*'.
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| '?' Matches any single character.
'*' Matches any sequence of characters (including the empty sequence).
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The matching should cover the entire input string (not partial).
Note:
s could be empty and contains only lowercase letters a-z.p could be empty and contains only lowercase letters a-z, and characters like ? or *.
Example 1:
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| Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".
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Example 2:
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| Input:
s = "aa"
p = "*"
Output: true
Explanation: '*' matches any sequence.
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Example 3:
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| Input:
s = "cb"
p = "?a"
Output: false
Explanation: '?' matches 'c', but the second letter is 'a', which does not match 'b'.
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Example 4:
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| Input:
s = "adceb"
p = "*a*b"
Output: true
Explanation: The first '*' matches the empty sequence, while the second '*' matches the substring "dce".
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Example 5:
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| Input:
s = "acdcb"
p = "a*c?b"
Output: false
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1 动态规划1
和leetcode 10 正则表达式匹配一样, 采用动态规划
思路可参考下面的URL
https://leetcode-cn.com/problems/wildcard-matching/solution/tong-pei-fu-pi-pei-by-leetcode-solution/
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| class Solution {
public boolean isMatch(String s, String p) {
final int m = s.length() + 1;
final int n = p.length() + 1;
boolean[][] dp = new boolean[m][n];
dp[0][0] = true;
for(int i = 1; i < m; ++i)
dp[i][0] = false;
for(int j = 1; j < n; ++j)
{
if(p.charAt(j-1) == '*')
dp[0][j] = dp[0][j-1];
else
dp[0][j] = false;
}
for(int i = 1; i < m; ++i)
{
for(int j = 1; j < n; ++j)
{
if(p.charAt(j-1) == '*')
{
dp[i][j] = dp[i-1][j] || dp[i][j-1];
}
else
{
if(p.charAt(j-1) == s.charAt(i-1) || p.charAt(j-1) == '?')
dp[i][j] = dp[i-1][j-1];
else
dp[i][j] = false;
}
}
}
return dp[m-1][n-1];
}
}
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2 动态规划2
发现动态规划1中, 数组dp[i]可以完全由dp[i-1]推导, 所以可以只new2个1维数组, dp_iMinus1[]对应dp[i-1][]和dp_i[]对应dp[i][]
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| class Solution {
public boolean isMatch(String s, String p) {
final int m = s.length() + 1;
final int n = p.length() + 1;
boolean[] dp_iMinus1 = new boolean[n];
boolean[] dp_i = new boolean[n];
dp_iMinus1[0] = true;
for(int j = 1; j < n; ++j)
{
if(p.charAt(j-1) == '*')
dp_iMinus1[j] = dp_iMinus1[j-1];
else
dp_iMinus1[j] = false;
}
for(int i = 1; i < m; ++i)
{
dp_i[0] = false;
for(int j = 1; j < n; ++j)
{
if(p.charAt(j-1) == '*')
{
dp_i[j] = dp_iMinus1[j] || dp_i[j-1];
}
else
{
if(p.charAt(j-1) == s.charAt(i-1) || p.charAt(j-1) == '?')
dp_i[j] = dp_iMinus1[j-1];
else
dp_i[j] = false;
}
}
dp_iMinus1 = Arrays.copyOf(dp_i, dp_i.length);
}
return dp_iMinus1[n-1];
}
}
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