583. Delete Operation for Two Strings

Given two strings word1 and word2, return the minimum number of steps required to make word1 and word2 the same.

In one step, you can delete exactly one character in either string.

题目解析：

• dp[i][j]：以i-1为结尾的字符串word1，和以j-1位结尾的字符串word2，想要达到相等，所需要删除元素的最少次数。
• 当word1[i - 1] 与 word2[j - 1]相同的时候，dp[i][j] = dp[i - 1][j - 1];
• 当word1[i - 1] 与 word2[j - 1]不相同的时候，递推公式：dp[i][j] = min(dp[i - 1][j] + 1, dp[i][j - 1] + 1);
• 初始化：dp[i][0]：word2为空字符串，以i-1为结尾的字符串word1要删除多少个元素，才能和word2相同呢，很明显dp[i][0] = i。

第二种方法就是求两个字符串的最长公共子序列
本题和动态规划：1143.最长公共子序列 (opens new window)基本相同，只要求出两个字符串的最长公共子序列长度即可，那么除了最长公共子序列之外的字符都是必须删除的，最后用两个字符串的总长度减去两个最长公共子序列的长度就是删除的最少步数。
word1.size()+word2.size()-dp[word1.size()][word2.size()]*2;

代码：

class Solution {
    public int minDistance(String word1, String word2) {
        int[][] dp = new int[word1.length() + 1][word2.length() + 1];
        dp[0][0] = 0;
        for (int i = 1; i <= word1.length(); i++) {
            dp[i][0] = i;
        }
        for (int j = 1; j <= word2.length(); j++) {
            dp[0][j] = j;
        }
        for (int i = 0; i < word1.length(); i++) {
            for (int j = 0; j < word2.length(); j++) {
                if (word1.charAt(i) == word2.charAt(j)) {
                    dp[i+1][j+1] = dp[i][j];
                } else {
                    dp[i+1][j+1] = Math.min(dp[i+1][j] + 1, dp[i][j+1] + 1);
                }
            }
        }
        return dp[word1.length()][word2.length()];
    }
}

72. Edit Distance

Given two strings word1 and word2, return the minimum number of operations required to convert word1 to word2.

You have the following three operations permitted on a word:

• Insert a character
• Delete a character
• Replace a character

题目解析：

• 与上题类似

代码：

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++) {
            dp[i][0] = i;
        }
        for (int j = 1; j <= word2.length(); j++) {
            dp[0][j] = j;
        }
        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];
                } else {
                    dp[i][j] = Math.min(dp[i-1][j], dp[i][j-1]);
                    dp[i][j] = Math.min(dp[i][j], dp[i-1][j-1]) + 1;
                }
            }
        }
        return dp[word1.length()][word2.length()];
    }
}