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…

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…

最小编辑距离,动态规划经典题. 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…

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 题意: 给定两个串串: word1, word2;  找…

Leetcode72 看起来比较棘手的一道题(列DP方程还是要大胆猜想..) DP方程该怎么列呢? dp[i][j]表示字符串a[0....i-1]转化为b[0....j-1]的最少距离 转移方程分三种情况考虑 分别对应三中操作 因为只需要三个值就可以更新dp[i][j] 我们可以把空间复杂度降低到O(n) Replace word1[i - 1] by word2[j - 1] (dp[i][j] = dp[i - 1][j - 1] + 1 (for replacement)); Delet…

原题 dp 利用二维数组dp[i][j]存储状态: 从字符串A的0~i位子字符串 到 字符串B的0~j位子字符串,最少需要几步.(每一次删增改都算1步) 所以可得边界状态dp[i][0]=i,dp[0][j]=j. 以及状态转移方程 即当比较 word1[i] 和 word2[j] 字符 相等 时,所需步数与 word1[i-1] 和 word2[j-1] 相等. 状态转移方程为:dp[i][j]=dp[i-1][j-1] 否则,状态转移方程为dp[i][j]= min(dp[i-1][j-1]…

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…

一.题目说明 题目72. Edit Distance,计算将word1转换为word2最少需要的操作.操作包含:插入一个字符,删除一个字符,替换一个字符.本题难度为Hard! 二.我的解答 这个题目一点思路也没,就直接看答案了.用的还是dp算法,dp[n1+1][n2+1]中的dp[i][j]表示将word1的前i位,变为word2的前j位需要的步骤.注意第1行是空,第1列也是空. 1.第一行中,dp[0][i]表示空字符""到word2[0,...,i]需要编辑几次 2.第一列中,d…

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…