LeetCode-Edit Distance

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#leetcode #Edit Distance

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本文介绍了一种求解两字符串间转换所需的最小操作步骤的算法。通过插入、删除或替换字符实现转换，使用动态规划方法高效计算最短路径。

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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

Solution:

Code;

<span style="font-size:14px;">class Solution {
public:
    int minDistance(string word1, string word2) {
        int length1 = word1.size(), length2 = word2.size();
        int **dp = new int *[length1+1];
        for (int i = 0; i < length1+1; i++)
            dp[i] = new int[length2+1];
        dp[0][0] = 0;
        for (int i = 1; i < length1+1; i++)
            dp[i][0] = i;
        for (int i = 1; i < length2+1; i++)
            dp[0][i] = i;
        for (int i = 1; i < length1+1; i++)
            for (int j = 1; j < length2+1; j++)
                if (word1[i-1] == word2[j-1]) {
                    dp[i][j] = min(dp[i][j-1]+1, dp[i-1][j]+1);
                    dp[i][j] = min(dp[i-1][j-1], dp[i][j]);
                } else {
                    dp[i][j] = min(dp[i][j-1]+1, dp[i-1][j]+1);
                    dp[i][j] = min(dp[i-1][j-1]+1, dp[i][j]);
                }
         int result = dp[length1][length2];
         for (int i = 0; i < length1+1; i++)
            delete [] dp[i];
        delete [] dp;
        return result;
    }
};</span>