1143. 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.

Longest Common Subsequence (LCS)的題目可以用DP來做。dp[i][j]代表在text1[0...i - 1]和text2[0...j - 1]中的LCS長度。 dp state transition: dp[i][j] = dp[i - 1][j - 1] + 1 if text1[i - 1] == text2[j - 1] dp[i][j] = max(dp[i][j - 1], dp[i - 1][j]) if text1[i - 1] != text2[j - 1]

// Bottom-Up Dynamic Programming
int longestCommonSubsequence(string text1, string text2) { // time: O(m * n); space: O(m * n)
    int m = text1.length(), n = text2.length();
    vector<vector<int> > dp(m + 1, vector<int> (n + 1, 0));
    for (int i = 1; i <= m; ++i) {
        for (int j = 1; j <= n; ++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[m][n];
}

5. Longest Palindromic Substring