[LeetCode] 583. Delete Operation for Two Strings 兩個字符串的刪除操做

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.

求出最長相同子序列Longest Common Subsequence，而後兩個單詞長度和減去2倍的相同子序列長度就是答案。

解法1: 遞歸， 若是[0, i], [0, j]最後一個字符相同，則比較[0, i-1], [0, j-1]的最後一個字符，若不相同，則刪去第i個或第j個字符後，返回長度更長的子序列。TLE

解法2：動態規劃dp，dp[i][j]表示word1的前i個字符和word2的前j個字符組成的兩個單詞的最長公共子序列的長度。若是當前的兩個字符相等，那麼dp[i][j] = dp[i-1][j-1] + 1 , 假設[0,i],[0,j]的最後一個字符匹配，則LCS的長度取決於第i-1和j-1個字符；若是不匹配，則須要進行錯位比較，也就是說，LCS的長度取決於[i-1]或[j-1]（取較長的一個）

Java1:

public class Solution {
    public int minDistance(String s1, String s2) {
        return s1.length() + s2.length() - 2 * lcs(s1, s2, s1.length(), s2.length());
    }
    public int lcs(String s1, String s2, int m, int n) {
        if (m == 0 || n == 0)
            return 0;
        if (s1.charAt(m - 1) == s2.charAt(n - 1))
            return 1 + lcs(s1, s2, m - 1, n - 1);
        else
            return Math.max(lcs(s1, s2, m, n - 1), lcs(s1, s2, m - 1, n));
    }
}

Java2:

class Solution {
    public int minDistance(String word1, String word2) {
        int dp[][]=new int[word1.length()+1][word2.length()+1];
        for(int i=0;i<word1.length()+1;++i){
            for(int j=0;j<word2.length()+1;++j){
                if(i==0||j==0)
                    continue;
                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]);
            }
        }
        return word1.length()+word2.length()-2*dp[word1.length()][word2.length()];
    }
}

Java:

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

Python1:

class Solution(object):
    def minDistance(self, word1, word2):
        """
        :type word1: str
        :type word2: str
        :rtype: int
        """
        return len(word1) + len(word2) - 2 * self.lcs(word1, word2)

    def lcs(self, word1, word2):
        len1, len2 = len(word1), len(word2)
        dp = [[0] * (len2 + 1) for x in range(len1 + 1)]
        for x in range(len1):
            for y in range(len2):
                dp[x + 1][y + 1] = max(dp[x][y + 1], dp[x + 1][y])
                if word1[x] == word2[y]:
                    dp[x + 1][y + 1] = dp[x][y] + 1
        return dp[len1][len2]　　

Python2:

class Solution(object):
    def minDistance(self, word1, word2):
        """
        :type word1: str
        :type word2: str
        :rtype: int
        """
        m, n = len(word1), len(word2)
        dp = [[0] * (n+1) for _ in xrange(2)]
        for i in xrange(m):
            for j in xrange(n):
                dp[(i+1)%2][j+1] = max(dp[i%2][j+1], \
                                       dp[(i+1)%2][j], \
                                       dp[i%2][j] + (word1[i] == word2[j]))
        return m + n - 2*dp[m%2][n]

C++:

class Solution {
public:
    int minDistance(string word1, string word2) {
        int n1 = word1.size(), n2 = word2.size();
        vector<vector<int>> dp(n1 + 1, vector<int>(n2 + 1, 0));
        for (int i = 1; i <= n1; ++i) {
            for (int j = 1; j <= n2; ++j) {
                if (word1[i - 1] == word2[j - 1]) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                } else {
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        return n1 + n2 - 2 * dp[n1][n2];
    }
};

C++:

class Solution {
public:
    int minDistance(string word1, string word2) {
        int n1 = word1.size(), n2 = word2.size();
        vector<vector<int>> dp(n1 + 1, vector<int>(n2 + 1, 0));
        for (int i = 0; i <= n1; ++i) dp[i][0] = i;
        for (int j = 0; j <= n2; ++j) dp[0][j] = j;
        for (int i = 1; i <= n1; ++i) {
            for (int j = 1; j <= n2; ++j) {
                if (word1[i - 1] == word2[j - 1]) {
                    dp[i][j] = dp[i - 1][j - 1];
                } else {
                    dp[i][j] = 1 + min(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        return dp[n1][n2];
    }
};

　　

　

相似題目：

[LeetCode] 712. Minimum ASCII Delete Sum for Two Strings 兩個字符串的最小ASCII刪除和

 

All LeetCode Questions List 題目彙總