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];
  }
};