在一个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?
如输入这样的一个二维数组,
[
[1,3,1],
[1,5,1],
[4,2,1]
]
[
[1,3,1],
[1,5,1],
[4,2,1]
]
那么路径 1→3→5→2→1 可以拿到最多价值的礼物,价值为12
[[1,3,1],[1,5,1],[4,2,1]]
12
# # 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 # class Solution: def maxValue(self, grid: List[List[int]]) -> int: # write code here M = len(grid) N = len(grid[0]) dp = [[0]*N for m in range(M)] # 记录当前位置的礼物的最大 for i in range(0, len(grid)): for j in range(0, len(grid[0])): # 初始化第一行,第一列的元素,它不可能从上面,或者左边移动过来,所以说它的大小就是它最大的价值 if i == 0 and j == 0: print(grid[0][0]) dp[i][j] = grid[i][j] # 棋盘中非第一行,非第一列中的元素的最大的情况 elif len(grid) > i > 0 and len(grid[0]) > j > 0: dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]) + grid[i][j] # 棋盘的第一行只能从左移动过去,从上面移动下来不可能 elif i-1 < 0: print("行列", i, j) dp[i][j] = dp[i][j-1]+grid[i][j] # 棋盘的第一列,不可能从左边移动过来,所以只能从上面移动过来 elif j-1 < 0: dp[i][j] = dp[i-1][j] + grid[i][j] return dp[M-1][N-1]
class Solution: # 动态规划 def maxValue(self , grid: List[List[int]]) -> int: rows = len(grid) cols = len(grid[0]) # 第一列只能来自上方 for i in range(1, rows): grid[i][0] += grid[i- 1][0] # 第一行只能来自左边 for i in range(1, cols): grid[0][i] += grid[0][i -1] # 遍历后续每一个位置 for i in range(1, rows): for j in range(1, cols): # 增加来自左边与上边的较大值 grid[i][j] += max(grid[i -1][j], grid[i][j -1]) return grid[rows -1][cols -1]
class Solution: def maxValue(self , grid): m,n=len(grid),len(grid[0]) dp=[[0]*n for _ in range(m)] dp[0][0]=grid[0][0] for i in range(m): for j in range(n): if m==0 and n==0: continue x= dp[i-1][j] if 0<=i-1<m else 0 y= dp[i][j-1] if 0<=j-1<n else 0 dp[i][j]=max(x,y)+grid[i][j] return dp[-1][-1]
class Solution: def maxValue(self , grid: List[List[int]]) -> int: m = len(grid) n = len(grid[0]) dp = [[0 for _ in range(n)] for _ in range(m)] dp[0][0] = grid[0][0] for i in range(1, m): dp[i][0] = dp[i - 1][0] + grid[i][0] for i in range(1, n): dp[0][i] = dp[0][i - 1] + grid[0][i] for i in range(1, m): for j in range(1, n): if i < m and j < n: dp[i][j] = max(dp[i][j - 1] + grid[i][j], dp[i - 1][j] + grid[i][j]) return dp[m - 1][n - 1]
class Solution: def maxValue(self , grid: List[List[int]]) -> int: # write code here m=len(grid) n=len(grid[0]) for i in range(m): for j in range(n): if i==0 and j==0: grid[i][j]=grid[i][j] elif i==0: grid[i][j]=grid[i][j-1]+grid[i][j] elif j==0: grid[i][j]=grid[i-1][j]+grid[i][j] else: grid[i][j]=max(grid[i-1][j],grid[i][j-1])+grid[i][j] return grid[-1][-1]
class Solution: def maxValue(self, grid): if not grid: return 0 dp = grid w, l = len(grid), len(grid[0]) for i in range(w): for j in range(l): # 第一个元素的特殊情况 if not (i&nbs***bsp;j): dp[i][j] = grid[i][j] elif i == 0: dp[i][j] = dp[i][j - 1] + grid[i][j] elif j == 0: dp[i][j] = dp[i - 1][j] + grid[i][j] else: dp[i][j] = max(dp[i-1][j], dp[i][j-1]) + grid[i][j] return dp[w-1][l-1]