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44. 通配符匹配 - 力扣（LeetCode）

输入：s = "aa", p = "a"
输出：false
解释："a" 无法匹配 "aa" 整个字符串。

输入：s = "aa", p = "*"
输出：true
解释：'*' 可以匹配任意字符串。

输入：s = "cb", p = "?a"
输出：false
解释：'?' 可以匹配 'c', 但第二个 'a' 无法匹配 'b'。

题解

题目描述：给你两个字符串s文本字符串和p模式字符串，其中p包含两种类型的通配符：

• *：可以匹配任意字符序列（包括空序列）
• ?：可以匹配任意单个字符

题目要求你检查s和p是否匹配。

Example 1:

• s = "adceb"
• p = "*a*b"
• Output: true
• Explanation: The first * can match an empty sequence, and the second * matches “dce”.

Example 2:

• s = "acdcb"
• p = "a*c?b"
• Output: false
• Explanation: There is no way to match s with p, as the character before the last in s is ‘c’ but ‘b’ in p.

这个题用贪心的思路该如何去想呢，在每一步匹配中，我们可以对如何匹配字符串做出局部最优选择，如*,可以表示任何字符序列，包括空序列，可以在匹配字符串的过程中及时调整匹配。

• 对于?：由于?只匹配一个字符，即自动选择s的当前字符与p中的?匹配。
• 对于*：因为他可以匹配任意字符序列，所以按照贪心的思路，首先用空序列匹配*，然后，如果需要，根据模式和字符串其余部分扩展或缩小匹配的字符数量。

贪心算法的实现思路

1. 两个指针+回溯：首先定义两个指针，分别是s（sIdx）和p（pIdx），进行模式匹配，如果出现不匹配，则回溯到p中*的最后一个位置（如果有），并尝试不同的匹配。
2. ?通配符处理：在迭代中，如果 s 和 p 中的当前字符匹配，或者 p[pIdx] 是 ? ，只需将两个指针向前移动。
3. 处理*通配符：如果p是*，需要分别标记*在p中的位置（用starIdx）,和在s中的对应位置(sTmpIdx表示)。表示*匹配s中的空字符的起点。如果稍后出现不匹配，则回溯到starIdx并增加sTmpIdx，表示*现在应该在s中多匹配一个字符。将pIdx重置为starIdx+1，将 sIdx 重置为 sTmpIdx ，继续匹配。
4. 最后检查：遍历 s 后，确保 p 中所有剩余的字符都是 * 。如果是，它们可以匹配空序列，因此，模式匹配字符串。

具体示例：

Consider ：s = "adceb" and p = "*a*b"

Initial Setup

• s = "adceb"
• p = "*a*b"
• sIdx = 0, pIdx = 0, starIdx = -1, sTmpIdx = -1

Iteration 1

• p[0] is *, so we update starIdx = 0 and sTmpIdx = 0.
• Increment pIdx to 1.
• sIdx remains 0.

Iteration 2

• p[1] is 'a', but s[0] is 'a'. This is a mismatch.
• Since starIdx != -1, we use the star to match one more character.
• Increment sTmpIdx to 1. So sTmpIdx = 1.
• Update pIdx = starIdx + 1, which means pIdx = 1.
• Update sIdx = sTmpIdx, which means sIdx = 1.

Iteration 3

• Now p[1] is 'a' and s[1] is also 'a'. This is a match.
• Increment both sIdx and pIdx by 1.
• sIdx = 2, pIdx = 2.

Iteration 4

• p[2] is '*'. Update starIdx = 2 and sTmpIdx = 2.
• Increment pIdx to 3.
• sIdx remains 2.

Iteration 5

• p[3] is 'b', but s[2] is 'd'. This is a mismatch.
• Since starIdx != -1, we use the star to match one more character.
• Increment sTmpIdx to 3. So sTmpIdx = 3.
• Update pIdx = starIdx + 1, which means pIdx = 3.
• Update sIdx = sTmpIdx, which means sIdx = 3.

Iteration 6

• p[3] is 'b' and s[3] is 'c'. This is a mismatch.
• We repeat the backtracking process:
• Increment sTmpIdx to 4. So sTmpIdx = 4.
• Update pIdx = starIdx + 1, which means pIdx = 3.
• Update sIdx = sTmpIdx, which means sIdx = 4.

Iteration 7

• p[3] is 'b' and s[4] is 'b'. This is a match.
• Increment both sIdx and pIdx by 1.
• sIdx = 5, pIdx = 4.

Final Check

• sIdx is now equal to s.size(), so we exit the while loop.
• Check remaining characters in p. Since pIdx is also at the end of p, there are no more characters to check.

class Solution {
public:bool isMatch(string s, string p) {int sIdx = 0, pIdx = 0;      // Pointers for s and p.int starIdx = -1, sTmpIdx = -1;  // Indices to track the most recent '*' position in p and the corresponding position in s.while (sIdx < s.size()) {// If the characters match or pattern has '?', move both pointers.if (pIdx < p.size() && (p[pIdx] == s[sIdx] || p[pIdx] == '?')) {++sIdx;++pIdx;}// If pattern has '*', record the position and assume it matches zero characters initially.else if (pIdx < p.size() && p[pIdx] == '*') {starIdx = pIdx;sTmpIdx = sIdx;++pIdx;}// If a mismatch occurs and there was a previous '*', backtrack.// Try to match '*' with one more character in s.else if (starIdx != -1) {pIdx = starIdx + 1;sIdx = ++sTmpIdx;}// If no '*' to backtrack to, return false.else {return false;}}// Check if the remaining characters in pattern are all '*'.// They can match the empty sequence at the end of s.for (; pIdx < p.size(); ++pIdx) {if (p[pIdx] != '*') {return false;}}// If we've processed both strings completely, it's a match.return true;}
};