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…

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…

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. 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…

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. 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…