56. Edit Distance && Simplify Path

最新推荐文章于 2025-08-13 10:25:05 发布

转载最新推荐文章于 2025-08-13 10:25:05 发布 · 53 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/liyangguang1988/p/3954340.html

文章标签：

#数据结构与算法

本文介绍了两种常见的字符串处理算法：编辑距离算法用于计算两个字符串之间的相似度；路径简化算法用于处理Unix风格的文件路径，使其更加简洁。通过动态规划解决编辑距离问题，并给出了一种有效简化路径的方法。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Edit Distance

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

a) Insert a character b) Delete a character c) Replace a character

思路：动态规划。
D[i+1][j+1] = D[i][j]; word1[i] == word2[j],
D[i+1][j+1] = min(min(D[i][j+1], D[i+1][j]), D[i][j]) + 1; otherwise.

class Solution {
public:
    int minDistance(string word1, string word2) {
        vector<vector<int> > D(word1.size()+1, vector<int>(word2.size()+1, 0));
        for(int j = 0; j <= word2.size(); ++j) 
            D[0][j] = j;
        for(int i = 0; i <= word1.size(); ++i) 
            D[i][0] = i;
        for(int i = 0; i < word1.size(); ++i) {
            for(int j = 0; j < word2.size(); ++j) {
                if(word1[i] == word2[j]) 
                    D[i+1][j+1] = D[i][j];
                else D[i+1][j+1] = min(min(D[i][j+1], D[i+1][j]), D[i][j]) + 1;
            }
        }
        return D[word1.size()][word2.size()];
    }
};

Simplify Path

Given an absolute path for a file (Unix-style), simplify it.

For example, path = "/home/", => "/home" path = "/a/./b/../../c/", => "/c"

click to show corner cases.

Corner Cases:

Did you consider the case where path = "/../"? In this case, you should return "/".
Another corner case is the path might contain multiple slashes '/' together, such as "/home//foo/". In this case, you should ignore redundant slashes and return "/home/foo".

注意： /..., /.home, /..h2me, /ho_Me/... 为合法路径。

思路：从头往尾读：如是 / 和字符，数字，下划线 , 好判断。若是 '.'，则分情况即可。

inline bool isAlpha(char ch) {
	return(('a' <= ch && ch <= 'z') || ('A' <= ch && ch <= 'Z'));
}
inline bool isAlphaOrUnderline(char ch) {
	return isAlpha(ch) || (ch == '_');
}
inline bool isValid(char ch) {
	return isAlphaOrUnderline(ch) || ('0' <= ch && ch <= '9');
}
class Solution {// the first alpha should be '/'
public:
	string simplifyPath(string path) {
		string ans;
		path.insert(0, 1, '/'); // but without this state. is OK too.
		for(size_t i = 0; i < path.size(); ++i) {
			if(path[i] == '.') {
				if(i < path.size()-1 && isAlphaOrUnderline(path[i+1])) { ans.push_back('.'); continue;}
				else if(i < path.size()-2 && path[i+1] == '.' && (isAlphaOrUnderline(path[i+2]) || path[i+2] == '.')) {
					i += 2;
					ans.insert(ans.size(), 2, '.');
					ans.push_back(path[i]);
					continue;
				}
			}
			if(path[i] == '/' && !ans.empty() && ans.back() == '/') continue;
			if('0' <= path[i] && path[i] <= '9' && !ans.empty() && ans.back() == '/') continue;
			if(path[i] == '/' || isValid(path[i])) ans.push_back(path[i]);
			else if(path[i] == '.' && i < path.size()-1 && path[i+1] == '.') {
				++i;
				if(ans.size() > 1 && ans.back() == '/') ans.pop_back();
				while(!ans.empty() && ans.back() != '/') ans.pop_back();
			}
		}
		if(ans.size() > 1 && ans.back() == '/')ans.pop_back(); 
		return ans;
	}
};

转载于:https://www.cnblogs.com/liyangguang1988/p/3954340.html