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 character b) Delete a character c) R…

这样的字符转换的dp挺经典的, 若word1[i+1]==word2[j+1] dp[i+1][j+1] = dp[i][j]:否则,dp[i+1][j+1] = dp[i][j] + 1.(替换原则),dp[i+1][j+1]还能够取dp[i][j+1]与dp[i+1][j]中的较小值.(删除加入原则) class Solution { public: int minDistance(string word1, string word2) { int dp[word1.size()+1][wo…

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…

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…

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,…

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. 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…

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…

引言 二维动态规划中最常见的是棋盘型二维动态规划. 即 func(i, j) 往往只和 func(i-1, j-1), func(i-1, j) 以及 func(i, j-1) 有关 这种情况下,时间复杂度 O(n*n),空间复杂度往往可以优化为O(n) 例题  1 Minimum Path Sum  Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right whi…

Edit Distance 题解 原创文章,拒绝转载 题目来源:https://leetcode.com/problems/edit-distance/description/ Description 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…

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…

Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2. You have the following 3 operations permitted on a word: Insert a character Delete a character Replace a character Example 1: Input: word1 = "h…

Leetcode之动态规划(DP)专题-72. 编辑距离(Edit Distance) 给定两个单词 word1 和 word2,计算出将 word1 转换成 word2 所使用的最少操作数 . 你可以对一个单词进行如下三种操作: 插入一个字符 删除一个字符 替换一个字符 示例 1: 输入: word1 = "horse", word2 = "ros" 输出: 3 解释: horse -> rorse (将 'h' 替换为 'r') rorse -> r…

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…

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…

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…

72. 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…

1. N-Queens The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other. Given an integer n, return all distinct solutions to the n-queens puzzle. Each solution contains a distinct board confi…

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 character b) Delete a character c) Re…

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…

传送门 Description 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 cha…

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.…

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…

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 {…

这里记录一下自己刷的LeetCode题目. 有些博客用英文阐述自己的思路和收获,相当于练习一下英文的表达能力. 比较好的题目有加粗. 1. Two Sum 3. Longest Substring Without Repeating Characters 5. Longest Palindromic Substring 7. Reverse Integer 9. Palindrome Number 8. String to Integer (atoi) 溢出的处理 21. Merge Two S…

本文已授权 [Coding博客](https://blog.coding.net) 转载 前言 Edit Distance,中文叫做编辑距离,在文本处理等领域是一个重要的问题,以下是摘自于百度百科的定义 编辑距离(Edit Distance),又称Levenshtein距离,是指两个字串之间,由一个转成另一个所需的最少编辑操作次数.许可的编辑操作包括将一个字符替换成另一个字符,插入一个字符,删除一个字符. 分别用R(replace),I(insert),D(delete),M(Match)代表替…

最小编辑距离的定义:编辑距离(Edit Distance),又称Levenshtein距离.是指两个字串之间,由一个转成还有一个所需的最少编辑操作次数.许可的编辑操作包含将一个字符替换成还有一个字符.插入一个字符,删除一个字符. 比如将kitten一字转成sitting: sitten(k→s) sittin(e→i) sitting(→g) 年提出这个概念. Thewords `computer' and `commuter' are very similar, and a change of…

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…