【Leetcode】583. Delete Operation for Two Strings

583. Delete Operation for Two Strings

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:

• The length of given words won't exceed 500.

• Characters in given words can only be lower-case letters.

题意

给两个单词，每次只能增删一个字符，从一个单词变到另一个需要多少步。

思路

实际上就是求两个字符串的相同部分，再用两个字符串的长度减去公共部分的长度，加和即为需要改变的次数。

本来我给弄成了最长公共子串，结果提交时发现WA了，看了测试用例才发现：

word1: park
word2: spake

结果应该为3，但是用最长公共子串的方式算出来为5，这里是因为应该使用最长公共子序列。

参考：最长公共子序列

代码

public class Solution {
	//最长公共子序列长度
    public int lcs(String s, String t) {
        if (s == null || s.length() == 0 || t == null || t.length() == 0) return 0;

        int len1 = s.length(), len2 = t.length();

        int[][] dp = new int[len1+1][len2+1];

        int max = 0;

        for (int i = 0; i <= len1; i++) {
            for (int j = 0; j <= len2; j++) {
                if (i==0||j==0) dp[i][j] = 0;
                else if (s.charAt(i-1) == t.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]);
                }

                if (max < dp[i][j])
                    max = dp[i][j];
            }
        }

        return max;

    }

    public int minDistance(String word1, String word2) {
        int common = lcs(word1, word2);
        return word2.length() + word1.length() - 2 * common;
    }
}