LeetCode1035. 不相交的线 (Java/C++ 动态规划)

Posted 2021-05-24 oyzg

Java/C++ 动态规划

链接：https://leetcode-cn.com/problems/uncrossed-lines/solution/javac-dong-tai-gui-hua-by-oyzg-tmyk/
解题思路：
1.dp[i][j]代表数组A[0…i]和数组B[0…j]的最大连线数
2.推导公式：
分2种情况：
（1）如果A数组和B数组的最后一个数相等，dp[i][j] = dp[i-1][j-1]+1;
（2）如果不相等，dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1]);

代码：
c++:

class Solution {
public:
    int maxUncrossedLines(vector<int>& A, vector<int>& B) {
		int n = A.size();
		 int m = B.size();
		 vector<vector<int>> dp(n + 1, vector<int>(m + 1));
		 for(int i = 1; i <= n; i++) {
			 for(int j = 1; j <= m; 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[n][m];
    }
};

Java:

class Solution {
    public int maxUncrossedLines(int[] A, int[] B) {
        int n = A.length;
		 int m = B.length;
		 int[][] dp = new int[n+1][m+1];
		 for(int i = 1; i <= n; i++) {
			 for(int j = 1; j <= m; j++) {
				 if(A[i-1] == B[j-1]) {
					 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[n][m];
    }
}