LeetCode 1143.最长公共子序列

LeetCode 1143.最长公共子序列

动态规划

1. 确定dp数组及其下标的含义

   dp[i] [j]:长度为**[0,i-1]的字符串text1与长度为[0,j-1]**的字符串text2的最长公共子序列为dp[i] [j]

2. 确定递推公式

   分两种情况：

   • text1[i-1]与text2[j-1]相同；即找到一个公共元素，dp[i] [j] = dp[i-1] [j-1] + 1;
   • text1[i-1]与text2[j-1]不同；找max(text1[0,i-2]与text2[0,j-1]和text1[0,i-1]与text2[0,j-2]的最长公共子序列),即 dp[i] [j] = max(dp[i - 1] [j], dp[i] [j - 1]);

   // 完整代码
   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]);
   }
   

3. dp数组初始化

   dp[i] [0]:0

   dp[0] [j]:0

完整代码：

class Solution {
public:
    int longestCommonSubsequence(string text1, string text2) {
		int m = text1.size(), n = text2.size();
		vector> res(m + 1, vector(n + 1, 0));
		for (int i = 1; i <= m; i++)
		{
			for (int j = 1; j <= n; j++)
			{
				if (text1[i - 1] == text2[j - 1])
					res[i][j] = res[i - 1][j - 1] + 1;
				else
					res[i][j] = max(res[i - 1][j], res[i][j - 1]);
			}
		}
		return res[m][n];
	}
};