【Leetcode】72.編輯距離

題目

給定兩個單詞 word1 和 word2，計算出將 word1 轉換成 word2 所使用的最少操做數 。
你能夠對一個單詞進行以下三種操做：

插入一個字符
刪除一個字符
替換一個字符

示例 1:

輸入: word1 = "horse", word2 = "ros"
輸出: 3
解釋: 
horse -> rorse (將 'h' 替換爲 'r')
rorse -> rose (刪除 'r')
rose -> ros (刪除 'e')

示例 2:

輸入: word1 = "intention", word2 = "execution"
輸出: 5
解釋: 
intention -> inention (刪除 't')
inention -> enention (將 'i' 替換爲 'e')
enention -> exention (將 'n' 替換爲 'x')
exention -> exection (將 'n' 替換爲 'c')
exection -> execution (插入 'u')

題解

這個題目拿到題目基本就能想到DP，由於感受和咱們以前的爬樓梯啥的比較類似。這個題目比較爲hard主要是，狀態轉換比較複雜。

定義: dpi ， word1這個字符串的前i個 -> word1這個字符串的前j 個字符，所須要的最小的步數

那麼有如下幾種狀況

word1[i] == word2[j]

不須要作變化，那麼 dpi = dp[i-1,j-1]

word1[i] != word2[j]

咱們就須要動用上面那三種操做了：

• add = dp[i, j-1], 表明插入一個字符
• delete = dp[i-1, j]，表明刪除一個字符
• replace = dp[i-1, j-1]，表明替換一個字符
• dpi = 1 + min(add, delete, replace)

時間複雜度 o(m * n)

java

class Solution {
    public int minDistance(String word1, String word2) {
        int m = word1.length();
        int n = word2.length();
        int[][] dp = new int[m + 1][n + 1];
        // base 
        for (int i = 0; i <= m; i++) {
            dp[i][0] = i;
        }

        for (int j = 0; j <= n; j++) {
            dp[0][j] = j;
        }
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (word1.charAt(i) == word2.charAt(j)) {
                    dp[i + 1][j + 1] = dp[i][j];
                } else {
                    int add = dp[i][j + 1];
                    int delete = dp[i + 1][j];
                    int rep = dp[i][j];
                    dp[i + 1][j + 1] = Math.min(Math.min(add, delete), rep) + 1;
                }
            }
        }
        return dp[m][n];
    }
}

python

class Solution:
    def minDistance(self, word1, word2):
        """
        :type word1: str
        :type word2: str
        :rtype: int
        """
        m = len(word1)
        n = len(word2)
        dp = [[0] * (n + 1) for _ in range(m + 1)]
        for i in range(m + 1):
            dp[i][0] = i
        for i in range(n + 1):
            dp[0][i] = i
        for i in range(m):
            for j in range(n):
                if word1[i] == word2[j]:
                    dp[i + 1][j + 1] = dp[i][j]
                else:
                    add = dp[i][j + 1]
                    delete = dp[i + 1][j]
                    replace = dp[i][j]
                    dp[i + 1][j + 1] = min(add, delete, replace) + 1
        return dp[m][n]

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