72. Edit Distance

微微清露

于 2017-06-28 22:12:30 发布

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本文链接：https://blog.youkuaiyun.com/fengyuhe1001/article/details/73864562

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本文介绍了一个经典的字符串处理问题——编辑距离。通过一个简洁高效的动态规划算法实现，文章详细解释了如何将一个单词转换为另一个单词所需的最小操作步骤，包括插入、删除和替换字符等操作。

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72. Edit Distance

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) Replace a character

代码如下：

class Solution { 
public:
    int minDistance(string word1, string word2) {
        int m = word1.length(), n = word2.length();
        vector<int> cur(m + 1, 0);
        for (int i = 1; i <= m; i++)
            cur[i] = i;
        for (int j = 1; j <= n; j++) {
            int pre = cur[0];
            cur[0] = j;
            for (int i = 1; i <= m; i++) {
                int temp = cur[i];
                if (word1[i - 1] == word2[j - 1])
                    cur[i] = pre;
                else cur[i] = min(pre + 1, min(cur[i] + 1, cur[i - 1] + 1));
                pre = temp;
            }
        }
        return cur[m]; 
    }
};