[LeetCode] 583. Delete Operation for Two Strings

标签：out   step   span   元素   letter   ndis   color   cas   cee   

Given two words word1 and word2, find the minimum number of steps required to make word1 and word2 the same, where in each step you can delete one character in either string.

Example 1:

Input: "sea", "eat"
Output: 2
Explanation: You need one step to make "sea" to "ea" and another step to make "eat" to "ea".

Note:

1. The length of given words won‘t exceed 500.
2. Characters in given words can only be lower-case letters.

题意：给两个字符串，问最多走(删除一个元素)可以使得两个字符串相等

反过来去理解题，删几个元素，可以看作，a中与b中不同的，加上b中与a中不同的

也就是说，res = s1.length() + s2.length() - 2*common 这题就转化为了找两个字符串最长的公共子序列串。

class Solution {
    public int minDistance(String word1, String word2) {
        int[] dp[] = new int[word1.length() + 1][word2.length() + 1];

        for (int i = 1; i <= word1.length(); i++)
            for (int j = 1; j <= word2.length(); j++) {
                if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                } else
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
            }
        int com = dp[word1.length()][word2.length()];
        return word1.length() + word2.length() - 2 * com;
    }
}

[LeetCode] 583. Delete Operation for Two Strings

标签：out   step   span   元素   letter   ndis   color   cas   cee   

原文地址：https://www.cnblogs.com/Moriarty-cx/p/9911545.html