剑指 Offer 47. 礼物的最大价值 发表于 2022-04-25 1234567891011121314151617181920212223242526class Solution { public int maxValue(int[][] grid) { int m = grid.length, n = grid[0].length; // 定义 dp[i][j] 为到达该格子(含)时礼物的最大价值 int[][] dp = new int[m][n]; dp[0][0] = grid[0][0]; // 注意不要忘记初始化边界值 for (int j = 1; j < n; j++) { dp[0][j] = grid[0][j] + dp[0][j - 1]; // 注意不要忘记累加之前路径上的和 dp[0][j - 1] } for (int i = 1; i < m; i++) { dp[i][0] = grid[i][0] + dp[i - 1][0]; // ditto } 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]; }} Reference剑指 Offer 47. 礼物的最大价值