0072.编辑距离

https://leetcode.cn/problems/edit-distance/

方法一：动态规划

时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

func minDistance(word1 string, word2 string) int {
	dp, m, n := make([][]int, len(word1)+1), len(word1), len(word2)
	for i := 0; i <= m; i++ {
		dp[i] = make([]int, n+1)
	}
	for i := 1; i <= m; i++ {
		dp[i][0] = i
	}
	for i := 1; i <= n; i++ {
		dp[0][i] = i
	}
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			if word1[i] == word2[j] {
				dp[i+1][j+1] = dp[i][j]
			} else {
				dp[i+1][j+1] = min(dp[i][j+1], min(dp[i][j], dp[i+1][j])) + 1
			}
		}
	}
	return dp[m][n]
}