标签：style blog http color os io strong ar for
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
Leetcode:Edit Distance 字符串编辑距离
标签：style blog http color os io strong ar for
原文地址：http://www.cnblogs.com/wwwjieo0/p/3941241.html
