LeetCode 30天挑戰 Day-26

LeetCode 30 days Challenge - Day 26

本系列將對LeetCode新推出的30天算法挑戰進行總結記錄，旨在記錄學習成果、方便未來查閱，同時望爲廣大網友提供幫助。

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Longest Common Subsequence

Given two strings text1 and text2, return the length of their longest common subsequence.

A subsequence of a string is a new string generated from the original string with some characters(can be none) deleted without changing the relative order of the remaining characters. (eg, “ace” is a subsequence of “abcde” while “aec” is not). A common subsequence of two strings is a subsequence that is common to both strings.

If there is no common subsequence, return 0.

Example 1:

Input: text1 = "abcde", text2 = "ace" 
Output: 3  
Explanation: The longest common subsequence is "ace" and its length is 3.

Example 2:

Input: text1 = "abc", text2 = "abc"
Output: 3
Explanation: The longest common subsequence is "abc" and its length is 3.

Example 3:

Input: text1 = "abc", text2 = "def"
Output: 0
Explanation: There is no such common subsequence, so the result is 0.

Constraints:

• 1 <= text1.length <= 1000
• 1 <= text2.length <= 1000
• The input strings consist of lowercase English characters only.

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Solution

題目要求分析：給定兩個字符串，求其中最長公共子串的長度。

解法：

本題需要注意區分子序列、子串：subsequence是子序列，子序列中各元素不需要連續，substring是子串，子串中每個元素必須是連續的。

本題是經典的動態規劃題，維護一個二維動態規劃數組dp，其中：

1. i：1 - text1.length()；j：1 - text2.length()；
2. dp[i][j]：text1[0:i-1]子串以及text2[0:j-1]子串之間最長公共子串的長度。
3. 更新規則（對text1的每一個位置 - i-1，遍歷text2所有的位置 - j-1）：
   1. 當text1[i-1] == text2[j-1]：dp[i][j] = dp[i-1][j-1] + 1;
   2. 否則： dp[i][j] = max(dp[i-1][j], dp[i][j-1]);

最終結果即爲 dp[text1.length()][text2.length()]。

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int longestCommonSubsequence(string text1, string text2) {
    vector<vector<int>> dp(text1.length()+1, vector<int>(text2.length()+1, 0));
    for (int i = 1; i <= text1.length(); i++) {
        for (int j = 1; j <= text2.length(); j++) {
            if (text1[i-1] == text2[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 dp[text1.length()][text2.length()];
}

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傳送門：Longest Common Subsequence

2020/4 Karl