标签：point har sequence http interview mon opera pre string
Problem statement:
Given two strings, find the longest common subsequence (LCS).
Your code should return the length of LCS.
What‘s the definition of Longest Common Subsequence?
For "ABCD" and "EDCA", the LCS is "A" (or "D", "C"), return 1.
For "ABCD" and "EACB", the LCS is "AC", return 2.
Solution:
This is a DP problem for two sequences. Such as 72. Edit Distance and 583. Delete Operation for Two Strings.
The key points is also dp[i][j] means the LCS of first i chars in A and first j char in B, and return dp[m][n]
class Solution { public: /** * @param A, B: Two strings. * @return: The length of longest common subsequence of A and B. */ int longestCommonSubsequence(string A, string B) { // write your code here int m = A.size(); int n = B.size(); 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 (A[i - 1] == B[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]; } };
标签：point har sequence http interview mon opera pre string
原文地址：http://www.cnblogs.com/wdw828/p/6858799.html
