1143. Longest Common Subsequence

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# 題目敘述

Given two strings text1 and text2 , return the length of their longest common subsequence. If there is no common subsequence, return 0 .

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.

• For example, "ace" is a subsequence of "abcde" .

A common subsequence of two strings is a subsequence that is common to both strings.

# 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.

# 解題思路

# Solution

class Solution {

    public int longestCommonSubsequence(String text1, String text2) {

        int[][] DP = new int[text1.length()+1][text2.length()+1];

        for (int i = 1; i <= text1.length(); i++) {

            for (int j = 1; j <= text2.length(); j++) {

                if (text1.substring(i - 1, i).equals(text2.substring(j - 1, j))) {

                    DP[i][j] = DP[i - 1][j - 1] + 1;

                } else {

                    DP[i][j] = Math.max(DP[i - 1][j], DP[i][j - 1]);

                }

            }

        }

        return DP[text1.length()][text2.length()];

    }

}

#include <string>

#include <string.h>

using namespace std;

class Solution {

 public:

  int longestCommonSubsequence(string text1, string text2) {

    int len1 = text1.size();

    int len2 = text2.size();

    int DP[len1 + 1][len2 + 1];

    memset(DP, 0, sizeof DP);

    for (int i = 1; i <= len1; i++) {

      for (int j = 1; j <= len2; 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[len1][len2];

  }

};