Question 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: Insert a character Delete a character Repla…
Given two strings S and T, determine if they are both one edit distance apart. 这道题是之前那道Edit Distance的拓展,然而这道题并没有那道题难,这道题只让我们判断两个字符串的编辑距离是否为1,那么我们只需分下列三种情况来考虑就行了: 1. 两个字符串的长度之差大于1,那么直接返回False 2. 两个字符串的长度之差等于1,那么长的那个字符串去掉一个字符,剩下的应该和短的字符串相同 3. 两个字符串的长度之…
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 characterb) Delete a characterc) Replace…
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 characterb) Delete a chara…
LintCode 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: Insert a character Delete a c…
Problem Introduction The edit distinct between two strings is the minimum number of insertions, deletions and mismatches in an alignment of two strings. Problem Description Task.The goal of this problem is to implement the algorithm for computing the…
Given two strings S and T, determine if they are both one edit distance apart. 分析:https://segmentfault.com/a/1190000003906621 虽然我们可以用Edit Distance的解法,看distance是否为1,但Leetcode中会超时.这里我们可以利用只有一个不同的特点在O(N)时间内完成.如果两个字符串只有一个编辑距离,则只有两种情况: 两个字符串一样长的时候,说明有一个替换…
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 characterb) Delete a chara…
Given two strings S and T, determine if they are both one edit distance apart. 给定两个字符串,判断他们是否是一步变换得到的. 在这里需要注意几点: 1.不等于1的变换都要返回false(包括变换次数等于0). 2.还有很多细节需要注意. 方法如下: 1.直接判断:1)如果差值大于1,直接返回false. 2)如果长度相同,那么依次判断,是否只有一个字母不一样. 3)如果不一样,那么看是否是只是多出了一个字母. p…
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 char…
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) Repla…
原题链接在这里:https://leetcode.com/problems/one-edit-distance/ Given two strings S and T, determine if they are both one edit distance apart. 与Edit Distance类似. 若是长度相差大于1, return false. 若是长度相差等于1, 遇到不同char时, 长的那个向后挪一位. 若是长度相等, 遇到不同char时同时向后挪一位. 出了loop还没有返回,…
题目: 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 characterb) Delete a characterc) Rep…
题目: Given two strings S and T, determine if they are both one edit distance apart. 链接: http://leetcode.com/problems/one-edit-distance/ 题解: 求两个字符串是否只有1个Edit Distance. 看着这道题又想起了Edit Distance那道.不过这道题不需要用DP,只用设一个boolean变量hasEdited来逐字符判断就可以了.写法大都借鉴了曹神的代码.…
Problem: Given two strings S and T, determine if they are both one edit distance apart. General Analysis: This problem is not hard. However, to write out more efficient and elegant solution, we need to dive more deep to understand the logic behind it…
One Edit Distance Given two strings S and T, determine if they are both one edit distance apart. 分析: 编辑距离复杂度为O(MN),而本题显然不能用这么高的复杂度:首先,可以通过判断两个字符串是否等长来决定用增一位.减一位.替换一位这三种方法之一来使得两个字符串等同,如果都不行,就return false:然后同时遍历S和T,第一次遇到不匹配的,就用刚才判断出的方法拯救一下:第二次还遇到不匹配的,就…
最近因为工作需要,学习了NLP的相关知识,简单动手实现了一下计算Edit Distance的算法,就是计算一个字符串要变成另一个字符串需要的代价,这其中采用Levenshtein方式,即规定一个插入和一个删除的代价是1,一次替换的代价是2. 简单的逻辑: 对于长度为M的字符串X,长度为N的字符串Y, Initialization: D(i,0)=i D(0,j)=j Recurrence Relation: for each i=1...M for each j=1...N D(i,j)=Min…
Given two strings S and T, determine if they are both one edit distance apart. 给定两个字符串S和T,确定它们是否都是是一步变换得到的. class Solution { func isOneEditDistance(_ s:String, _ t:String) -> Bool{ var num:Int = min(s.count, t.count) ..<num { if s.count == t.count {…
Difficulty: Medium More:[目录]LeetCode Java实现 Description Given two strings S and T, determine if they are both one edit distance apart. Intuition 同时遍历比较S和T的字符,直至遇到不同的字符:如果S和T的字符个数相同时,跳过不同的那个字符,继续遍历:如果S和T的字符个数相差为1时,跳过较长的字符串的当天字符,继续遍历.如果剩下的字符都相等,那么返回tr…
Given two strings S and T, determine if they are both one edit distance apart. Example Given s = "aDb", t = "adb" return true 思维惯性造成上来就想call Edit Distance的算法 然后看需要改多少步 后来想想这个问题“One”很特殊 要好好利用 才发现简单的string compare就可以解决 最后判断前面的字符全部相等的情况,此…
Given two strings s and t, determine if they are both one edit distance apart. Note: There are 3 possiblities to satisify one edit distance apart: Insert a character into s to get t Delete a character from s to get t Replace a character of s to get t…
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 characterb) Delete a chara…
