IT博客汇
  • 首页
  • 精华
  • 技术
  • 设计
  • 资讯
  • 扯淡
  • 权利声明
  • 登录 注册

    [原]LeetCode -- Wildcard Matching

    csharp25发表于 2017-02-19 22:45:19
    love 0
    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:


    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];
        }
    }




沪ICP备19023445号-2号
友情链接