面试题 47. 礼物的最大价值

输入:
[
  [1,3,1],
  [1,5,1],
  [4,2,1]
]
输出: 12
解释: 路径 1→3→5→2→1 可以拿到最多价值的礼物

class Solution:
    def maxValue(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        dp = [[0] * (n + 1) for _ in range(m + 1)]
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]) + grid[i - 1][j - 1]
        return dp[m][n]

class Solution {
    public int maxValue(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        int[][] dp = new int[m + 1][n + 1];
        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 - 1][j - 1];
            }
        }
        return dp[m][n];
    }
}

class Solution {
public:
    int maxValue(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        vector<vector<int> > dp(m + 1, vector<int>(n + 1, 0));
        for (int i = 1; i < m + 1; ++i) {
            for (int j = 1; j < n + 1; ++j) {
                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]) + grid[i - 1][j - 1];
            }
        }
        return dp[m][n];
    }
};

/**
 * @param {number[][]} grid
 * @return {number}
 */
var maxValue = function (grid) {
  const m = grid.length;
  const n = grid[0].length;
  let dp = new Array(m + 1);
  for (let i = 0; i < m + 1; ++i) {
    dp[i] = new Array(n + 1).fill(0);
  }
  for (let i = 1; i < m + 1; ++i) {
    for (let j = 1; j < n + 1; ++j) {
      dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]) + grid[i - 1][j - 1];
    }
  }
  return dp[m][n];
};

func maxValue(grid [][]int) int {
    m, n := len(grid), len(grid[0])
    dp := make([][]int, m + 1)
    for i := 0; i < m + 1; i++ {
        dp[i] = make([]int, n + 1)
    }
    for i := 1; i < m + 1; i++ {
        for j := 1; j < n + 1; j++ {
            dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]) + grid[i - 1][j - 1]
        }
    }
    return dp[m][n]
}

func max(a, b int) int {
    if (a > b) {
        return a
    }
    return b
}