Leetcode:Edit Distance 字符串编辑距离

原题戳我

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

 

解析：

典型动态规划题.

设 distance[i,j]表示 word1[i]和word2[j]之间的编辑距离

1. 当word1[i] == word2[j]时，则 distance[i,j] = distance[i-1, j-1]

2. 当word1[i] 不等于 word2[j]时，分三种情况，分别对应三种操作

    

word1插入1个字符，即 distance[i, j] = distance[i, j - 1] + 1
word1删除1个字符，即 distance[i, j] = distance[i - 1, j] + 1 word1替换1个字符，即 distance[i, j] = distance[i - 1, j - 1] + 1

 

class Solution {
public:
    int minDistance(string word1, string word2) {
        if (word1.size() == 0) return word2.size();
        if (word2.size() == 0) return word1.size();
        
        int rowNum = word1.size() + 1;
        int colNum = word2.size() + 1;
        std::vector<std::vector<int>> distance(rowNum, std::vector<int>(colNum, 0));
        for (int i = 0; i < rowNum; ++i) {
            distance.at(i).at(0) = i;
        }
        for (int j = 0; j < colNum; ++j) {
            distance.at(0).at(j) = j;
        }
        
        for (int i = 1; i < rowNum; ++i) {
            for (int j = 1; j < colNum; ++j) {
                if (word1.at(i - 1) == word2.at(j - 1)) {
                    distance.at(i).at(j) = distance.at(i - 1).at(j - 1);
                } else {
                    int deleteCost = distance.at(i - 1).at(j) + 1;
                    int insertCost = distance.at(i).at(j - 1) + 1;
                    int replaceCost = distance.at(i - 1).at(j - 1) + 1;
                    distance.at(i).at(j) = std::min(deleteCost, std::min(insertCost, replaceCost));
                }
            }
        }
        return distance.at(word1.size()).at(word2.size());
    }
};

 

注意：

distance二维矩阵的设定，行和列都是对应字符串长度加1

distance二维矩阵的 0行 和 0列的初始化，边界情况