题解 | #礼物的最大价值#
礼物的最大价值
https://www.nowcoder.com/practice/2237b401eb9347d282310fc1c3adb134
import java.util.*; public class Solution { /** * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 * * * @param grid int整型二维数组 * @return int整型 */ public int maxValue (int[][] grid) { int m = grid.length, n = grid[0].length; int[][] dp = new int[m][n]; dp[0][0] = grid[0][0]; for (int i = 1; i < m; i++) { dp[i][0] = dp[i - 1][0] + grid[i][0]; } for (int i = 1; i < n; i++) { dp[0][i] = dp[0][i - 1] + grid[0][i]; } for (int i = 1; i < m; i++) { for (int j = 1; j < n; j++) { dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]) + grid[i][j]; } } return dp[m - 1][n - 1]; } public int maxValue1 (int[][] grid) { int[][] dp = new int[grid.length + 1][grid[0].length + 1]; for (int i = 1; i < dp.length ; i++) { for (int j = 1 ; j < dp[0].length ; j++) { dp[i][j] = Math.max(dp[i - 1][j] + grid[i - 1][j - 1], dp[i][j - 1] + grid[i - 1][j - 1]); } } return dp[grid.length][grid[0].length]; } }
法1:设置一个m*n的dp table,base case为最左列和最上行,分别累加。法2设置一个(m+1)*(n+1)的dp table,最左行和最右列初始化为0,不需要累加。