给出两个字符串,找到最长公共子序列(LCS),返回LCS的长度。
动态规划。大神July讲的很详细,请大家移步这里。
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));
for(int i=0; i<=m; i++){
for(int j=0; j<=n; j++){
if(i==0 || j==0){
dp[i][j] = 0;
}
else{
if (A[i-1] == B[j-1]){
dp[i][j] = dp[i-1][j-1] + 1;
}
else{
dp[i][j] = max(max(dp[i-1][j], dp[i][j-1]), dp[i-1][j-1]);
}
}
}
}
return dp[m][n];
}
};
string A = “bedaacbade”;
string B = “dccaeedbeb”;
dp数组如下: