Leetcode 72. Edit Distance_given two words word1 and word2, find the minimum -优快云博客

本文介绍LeetCode第72题编辑距离问题，使用动态规划解决两字符串之间的最小编辑距离，包括插入、删除及替换操作。

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Leetcode 72. Edit Distance

题目描述

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

输入：两个字符串word1, word2
输出：需要修改、删除、插入的次数

思路

动态规划：用L(i,j)表示word1[0:i]与word2[0:j]的编辑距离

for i = 0，1, 2, ..., m:
    L(i,0) = i
for j = 0，1, 2, ..., n:
    L(0,j) = j

for i = 0，1, 2, ..., m:
    for j = 0，1, 2, ..., n:
        L(i,j) = min(L(i-1,j)+1, L(i,j-1)+1, diff+L(i-1,j-1))

return L(m,n)

代码

class Solution {
public:
    int minDistance(string word1, string word2) {
        if(word2.length()==0||word1.length()==0) return max(word2.length(),word1.length());

        int L[word1.length()+1][word2.length()+1];

        for(int i =0;i<=word1.length();i++)
            L[i][0] = i;
        for(int j =0;j<=word2.length();j++)
            L[0][j] = j;

        for(int i = 1;i<=word1.length();i++){
            for(int j =1;j<=word2.length();j++){

                if(j>=1&&i>=1){
                    int diff = (word1[i-1]!=word2[j-1]);
                    L[i][j] = min(min(1+L[i-1][j],1+L[i][j-1]),diff+L[i-1][j-1]);
                }
            }
        }
        return L[word1.length()][word2.length()];


    }
};