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输入：word1 = "horse", word2 = "ros"
输出：3
解释：
horse -> rorse (将 'h' 替换为 'r')
rorse -> rose (删除 'r')
rose -> ros (删除 'e')

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输入：word1 = "intention", word2 = "execution"
输出：5
解释：
intention -> inention (删除 't')
inention -> enention (将 'i' 替换为 'e')
enention -> exention (将 'n' 替换为 'x')
exention -> exection (将 'n' 替换为 'c')
exection -> execution (插入 'u')

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class Solution {
public:
    /*
    ## 解题思路
    * 动态规划
      1. dp[i][j]: 表示word1[0i]和word2[0j]的最小编辑距离；
      2. word1[i] == word2[j]，w[i], w[j]相等，不用考虑，所以:
            dp[i][j] = dp[i-1][j-1]，
      3. word1[i] != word2[j]:
            dp[i][j] = min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1
            其中：dp[i-1][j]表示删除word1[i]操作；
                 dp[i][j-1]表示删除word2[j]操作；
                 dp[i-1][j-1]表示替换word1[i]为word2[j]字符操作, 即操作替换后word1[i] == word2[j]
      4. 初始条件：
           dp[i][0] = i:
           dp[0][j] = j;
    */
    int minDistance(string word1, string word2) {
        int m = word1.size();
        int n = word2.size();
        vector<vector<int>> dp(m+1, vector<int>(n+1, 0));

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

        return dp[m][n];
    }
};

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// @lc code=start
impl Solution {
    /// ## 解题思路
    /// - 动态规划
    /// 1. 设 dp[i][j]: 表示word1[0i]和word2[0j]的最小编辑距离;
    /// 2. 如果 word1[i] == word2[j], 则不用编辑, 有:
    ///    dp[i][j] = dp[i-1][j-1]
    /// 3. 如果 word1[i] != word2[j], 则
    ///    dp[i][j] = min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1
    ///    其中:
    ///     - dp[i-1][j]: 删除word1[i];
    ///     - dp[i][j-1]: 删除word2[j];
    ///     - dp[i-1][j-1]:
    ///  4. 初始条件:
    ///     dp[i][0] = i
    ///     dp[0][j] = j
    pub fn min_distance(word1: String, word2: String) -> i32 {
        let mut dp = vec![vec![0_i32; word2.len() + 1]; word1.len() + 1];
        for i in 1..=word1.len() {
            dp[i][0] = i as i32;
        }
        for j in 1..=word2.len() {
            dp[0][j] = j as i32;
        }
        for (i, w1) in word1.bytes().enumerate() {
            for (j, w2) in word2.bytes().enumerate() {
                if w1 == w2 {
                    dp[i + 1][j + 1] = dp[i][j];
                } else {
                    dp[i + 1][j + 1] =
                        std::cmp::min(std::cmp::min(dp[i][j + 1], dp[i + 1][j]), dp[i][j]) + 1;
                }
            }
        }
        dp[word1.len()][word2.len()]
    }
}
// @lc code=end

struct Solution;