【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 dpij = \begin{cases} i & j = 0 \quad \text{(初始化)} \\ j & i = 0 \quad \text{(初始化)} \\ dpi-1j-1 & word1i-1 = word2j-1 \quad \\ \min(dpi-1j, dpij-1, dpi-1j-1) + 1 & word1i-1 \neq word2j-1 \end{cases} dpij=⎩ ⎨ ⎧ijdpi−1j−1min(dpi−1j,dpij−1,dpi−1j−1)+1j=0(初始化)i=0(初始化)word1i−1=word2j−1word1i−1=word2j−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