72. Edit Distance

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本文介绍了一个经典的字符串处理问题——编辑距离算法。通过动态规划的方法，实现了将一个字符串转化为另一个字符串所需的最少操作步骤，包括插入、删除和替换字符等操作。

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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

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题意：计算编辑距离

思路：dp[i][j]表示转换word1[0, i]到word2[0, j]所需的最小步数，则dp[i][j] = min{ dp[i-1][j-1] + carry, dp[i-1][j]+1, dp[i][j-1]+1} carry=0 if word1[i]==word2[j], else carry=1;

class Solution {
public:
	int minDistance(string word1, string word2) {
		int m = word1.size();
		int n = word2.size();
		if (m == 0 || n == 0)
			return m | n;
		vector<vector<int> > dp(m + 1, vector<int>(n + 1, 0));
		for (int i = 0; i < n + 1; i++){
			dp[0][i] = i;
		}
		for (int i = 0; i < m + 1; i++){
			dp[i][0] = i;
		}

		for (int i = 1; i < m + 1; i++){
			for (int j = 1; j < n + 1; j++){
				if (word1[i - 1] == word2[j - 1])
					dp[i][j] = dp[i - 1][j - 1];
				else
					dp[i][j] = min(dp[i - 1][j - 1], min(dp[i - 1][j], dp[i][j - 1])) + 1;
			}
		}
		return dp[m][n];
	}
};