【LeetCode热题100（90/100）】编辑距离

题目地址： 链接

状态转移方程：
d p [ i ] [ j ] = { i j = 0 (初始化) j i = 0 (初始化) d p [ i − 1 ] [ j − 1 ] w o r d 1 [ i − 1 ] = w o r d 2 [ j − 1 ] min ⁡ ( d p [ i − 1 ] [ j ] , d p [ i ] [ j − 1 ] , d p [ i − 1 ] [ j − 1 ] ) + 1 w o r d 1 [ i − 1 ] ≠ w o r d 2 [ j − 1 ] dp[i][j] = \begin{cases} i & j = 0 \quad \text{(初始化)} \\ j & i = 0 \quad \text{(初始化)} \\ dp[i-1][j-1] & word1[i-1] = word2[j-1] \quad \\ \min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1 & word1[i-1] \neq word2[j-1] \end{cases} dp[i][j]=⎩ ⎨ ⎧ijdp[i−1][j−1]min(dp[i−1][j],dp[i][j−1],dp[i−1][j−1])+1j=0(初始化)i=0(初始化)word1[i−1]=word2[j−1]word1[i−1]=word2[j−1]

/*
 * @lc app=leetcode.cn id=72 lang=typescript
 *
 * [72] 编辑距离
 */

// @lc code=start
function minDistance(word1: string, word2: string): number {
    const [n, m] = [word1.length, word2.length];
    const dp = Array.from({length: n + 1}, () => new Array(m + 1).fill(0));
    
    if(n == 0 || m == 0) return n + m;

    for(let i = 0; i < n; i ++) dp[i + 1][0] = i + 1;
    for(let j = 0; j < m; j ++) dp[0][j + 1] = j + 1;
    
    for(let i = 0; i < n; i ++) {
        for(let j = 0; j < m; j ++) {
            if(word1[i] == word2[j]) {
                dp[i + 1][j + 1] = dp[i][j];
            } else {
                dp[i + 1][j + 1] = Math.min(dp[i][j + 1], dp[i + 1][j], dp[i][j]) + 1;
            }
        }
    }
    return dp[n][m];
};
// case
// ""a"\n"aa""
// @lc code=end