class Solution {
public:
int maxValue(vector<vector<int>>& grid) {
}
};
剑指 Offer 47. 礼物的最大价值
在一个 m*n 的棋盘的每一格都放有一个礼物，每个礼物都有一定的价值（价值大于 0）。你可以从棋盘的左上角开始拿格子里的礼物，并每次向右或者向下移动一格、直到到达棋盘的右下角。给定一个棋盘及其上面的礼物的价值，请计算你最多能拿到多少价值的礼物？
示例 1:
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
提示：
0 < grid.length <= 2000 < grid[0].length <= 200原站题解
golang 解法, 执行用时: 16 ms, 内存消耗: 3.7 MB, 提交时间: 2023-03-08 09:21:57
func maxValue(grid [][]int) int {
n := len(grid[0])
f := grid[0]
for j := 1; j < n; j++ {
f[j] += f[j-1]
}
for _, row := range grid[1:] {
f[0] += row[0]
for j := 1; j < n; j++ {
f[j] = max(f[j-1], f[j]) + row[j]
}
}
return f[n-1]
}
func max(a, b int) int { if a < b { return b }; return a }
java 解法, 执行用时: 2 ms, 内存消耗: 43.9 MB, 提交时间: 2023-03-08 09:21:36
class Solution {
public int maxValue(int[][] grid) {
int m = grid.length, n = grid[0].length;
var f = grid[0]; // 这里没有拷贝，f 和 grid[0] 都持有同一段内存
for (int j = 1; j < n; ++j)
f[j] += f[j - 1];
for (int i = 1; i < m; ++i) {
f[0] += grid[i][0];
for (int j = 1; j < n; ++j)
f[j] = Math.max(f[j - 1], f[j]) + grid[i][j];
}
return f[n - 1];
}
}
python3 解法, 执行用时: 52 ms, 内存消耗: 16.1 MB, 提交时间: 2023-03-08 09:21:12
class Solution:
def maxValue(self, grid: List[List[int]]) -> int:
m, n = len(grid), len(grid[0])
f = grid[0] # 这里没有拷贝，f 和 grid[0] 都持有同一段内存
for j in range(1, n):
f[j] += f[j - 1]
for i in range(1, m):
f[0] += grid[i][0]
for j in range(1, n):
f[j] = max(f[j - 1], f[j]) + grid[i][j]
return f[-1]
java 解法, 执行用时: 1 ms, 内存消耗: 43.9 MB, 提交时间: 2023-03-08 09:19:59
class Solution {
private int[][] grid, memo;
public int maxValue(int[][] grid) {
this.grid = grid;
int m = grid.length, n = grid[0].length;
memo = new int[m][n];
return dfs(grid.length - 1, grid[0].length - 1);
}
private int dfs(int i, int j) {
if (i < 0 || j < 0)
return 0;
if (memo[i][j] > 0) // grid[i][j] 都是正数，记忆化的 memo[i][j] 必然为正数
return memo[i][j];
return memo[i][j] = Math.max(dfs(i, j - 1), dfs(i - 1, j)) + grid[i][j];
}
}
golang 解法, 执行用时: 12 ms, 内存消耗: 4.3 MB, 提交时间: 2023-03-08 09:19:36
// 记忆化搜索
func maxValue(grid [][]int) int {
m, n := len(grid), len(grid[0])
memo := make([][]int, m)
for i := range memo {
memo[i] = make([]int, n)
}
var dfs func(int, int) int
dfs = func(i, j int) int {
if i < 0 || j < 0 {
return 0
}
p := &memo[i][j]
if *p > 0 { // grid[i][j] 都是正数，记忆化的 memo[i][j] 必然为正数
return *p
}
*p = max(dfs(i, j-1), dfs(i-1, j)) + grid[i][j]
return *p
}
return dfs(len(grid)-1, len(grid[0])-1)
}
func max(a, b int) int { if a < b { return b }; return a }
python3 解法, 执行用时: 60 ms, 内存消耗: 21.6 MB, 提交时间: 2023-03-08 09:18:54
# 记忆化搜索
class Solution:
def maxValue(self, grid: List[List[int]]) -> int:
@cache
def dfs(i: int, j: int) -> int:
if i < 0 or j < 0:
return 0
return max(dfs(i, j - 1), dfs(i - 1, j)) + grid[i][j]
return dfs(len(grid) - 1, len(grid[0]) - 1)
java 解法, 执行用时: 2 ms, 内存消耗: 43.8 MB, 提交时间: 2023-03-08 09:16:06
class Solution {
public int maxValue(int[][] grid) {
int m = grid.length, n = grid[0].length;
int[][] f = new int[m + 1][n + 1];
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]) + grid[i - 1][j - 1];
}
}
return f[m][n];
}
}
golang 解法, 执行用时: 4 ms, 内存消耗: 3.7 MB, 提交时间: 2023-03-08 09:15:03
func maxValue(grid [][]int) int {
m, n := len(grid), len(grid[0])
for i := 0; i < m; i++ {
for j := 0; j < n; j++ {
if i == 0 && j == 0 {
continue
}
if i == 0 {
grid[i][j] += grid[i][j-1]
} else if j == 0 {
grid[i][j] += grid[i-1][j]
} else {
grid[i][j] += max(grid[i][j-1], grid[i-1][j])
}
}
}
return grid[m-1][n-1]
}
func max(x, y int) int { if x < y { return y }; return x }
python3 解法, 执行用时: 52 ms, 内存消耗: 16.3 MB, 提交时间: 2022-11-13 12:41:40
class Solution:
def maxValue(self, grid: List[List[int]]) -> int:
for i in range(len(grid)):
for j in range(len(grid[0])):
if i == 0 and j == 0: continue
if i == 0: grid[i][j] += grid[i][j - 1] # 0行上
elif j == 0: grid[i][j] += grid[i - 1][j] # 0列上
else: grid[i][j] += max(grid[i][j - 1], grid[i - 1][j])
return grid[-1][-1]