LeetCode | EditDistance

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

动态规划经典题目，找到递推关系，初始化，然后递推。

这里的dp[i][j]表示word1.substring(0, i+1)和word2.substring(0, j+1)的最小编辑距离。

代码

public class EditDistance {
	public int minDistance(String word1, String word2) {
		char[] word1CharArray = word1.toCharArray();
		char[] word2CharArray = word2.toCharArray();
		int M = word1CharArray.length;
		int N = word2CharArray.length;
		if (M == 0 || N == 0) {
			return M + N;
		}

		int[][] dp = new int[M][N];
		// init
		dp[0][0] = word1CharArray[0] == word2CharArray[0] ? 0 : 1;
		for (int i = 1; i < M; ++i) {
			dp[i][0] = word1CharArray[i] == word2CharArray[0] ? i
					: dp[i - 1][0] + 1;
		}
		for (int j = 1; j < N; ++j) {
			dp[0][j] = word1CharArray[0] == word2CharArray[j] ? j
					: dp[0][j - 1] + 1;
		}

		// dp
		for (int i = 1; i < M; ++i) {
			for (int j = 1; j < N; ++j) {
				if (word1CharArray[i] == word2CharArray[j]) {
					dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + 1;
					dp[i][j] = Math.min(dp[i][j], dp[i - 1][j - 1]);
				} else {
					dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]);
					dp[i][j] = Math.min(dp[i][j], dp[i - 1][j - 1]) + 1;
				}
			}
		}
		return dp[M - 1][N - 1];
	}
}

LeetCode | EditDistance,布布扣,bubuko.com

LeetCode | EditDistance

原文：http://blog.csdn.net/perfect8886/article/details/21565173