Commit 1f50fa2

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ruislan

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solved q583

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	`1`	`+use crate::q::Solution;`
	`2`	`+`
	`3`	`+#[allow(unused)]`
	`4`	`+impl Solution {`
	`5`	`+ pub fn min_distance(word1: String, word2: String) -> i32 {`
	`6`	`+ // 方法1`
	`7`	`+ // 这个题只用了删除,而没有替换或者插入操作`
	`8`	`+ // 那么也就是意味着,我们只需要找出最长公共子序列(LCS)就可以了`
	`9`	`+ // 这样保证双方保留的字符是最多的剩下的就是删除的`
	`10`	`+ // 不过,如果有替换和插入操作,那么我们就需要使用编辑距离的解法了(Edit distance)`
	`11`	`+ // AC 4ms 4.1mb 1306/1306`
	`12`	`+ let n = word1.len();`
	`13`	`+ let m = word2.len();`
	`14`	`+ let word1: Vec<char> = word1.chars().collect();`
	`15`	`+ let word2: Vec<char> = word2.chars().collect();`
	`16`	`+ let mut dp = vec![vec![0; m + 1]; n + 1];`
	`17`	`+ for i in 1..=n {`
	`18`	`+ for j in 1..=m {`
	`19`	`+ if word1[i - 1] == word2[j - 1] {`
	`20`	`+ dp[i][j] = dp[i - 1][j - 1] + 1;`
	`21`	`+ } else {`
	`22`	`+ dp[i][j] = dp[i - 1][j].max(dp[i][j - 1]);`
	`23`	`+ }`
	`24`	`+ }`
	`25`	`+ }`
	`26`	`+ (n + m - dp[n][m] * 2) as i32`
	`27`	`+ }`
	`28`	`+}`

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