剑指 Offer 47. 礼物的最大价值(动态规划)
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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 <= 200
0 < grid[0].length <= 200
方法一:常规动态规划时间和空间复杂度均为O(m * n)
1 class Solution { 2 public: 3 // dp[i][j]到达(i,j)方格时拿到礼物最多价值 4 // 状态转移方程dp[i][j] = grid[i][j] + max(dp[i][j - 1], dp[i - 1][j]) 5 int maxValue(vector<vector<int>>& grid) { 6 if (grid.size() == 0 || grid[0].size() == 0) { 7 return 0; 8 } 9 int row = grid.size(); 10 int col = grid[0].size(); 11 vector<vector<int>> dp(row, vector<int>(col, 0)); 12 dp[0][0] = grid[0][0]; 13 // 当i == 0 && j != 0时,dp[i][j] = dp[i][j - 1] + grid[i][j] 14 // 当i != 0 && j == 0时,dp[i][j] = dp[i - 1][j] + grid[i][j] 15 // 当i != 0 && j != 0时,dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]) + grid[i][j] 16 for (int i = 0; i < row; i++ ) { 17 for (int j = 0; j < col; j++) { 18 if (i == 0 && j == 0) { 19 continue; 20 } 21 if (i == 0 && j != 0) { 22 dp[i][j] = dp[i][j - 1] + grid[i][j]; 23 } else if (i != 0 && j == 0) { 24 dp[i][j] = dp[i - 1][j] + grid[i][j]; 25 } else { 26 dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]) + grid[i][j]; 27 } 28 } 29 } 30 return dp[row - 1][col - 1]; 31 } 32 };
方法二:优化空间复杂度为O(1)后的动态规划
1 class Solution { 2 public: 3 // dp[i][j]到达(i,j)方格时拿到礼物最多价值 4 // 状态转移方程dp[i][j] = grid[i][j] + max(dp[i][j - 1], dp[i - 1][j]) 5 int maxValue(vector<vector<int>>& grid) { 6 if (grid.size() == 0 || grid[0].size() == 0) { 7 return 0; 8 } 9 int row = grid.size(); 10 int col = grid[0].size(); 11 // 当i == 0 && j != 0时,dp[i][j] = dp[i][j - 1] + grid[i][j] 12 // 当i != 0 && j == 0时,dp[i][j] = dp[i - 1][j] + grid[i][j] 13 // 当i != 0 && j != 0时,dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]) + grid[i][j] 14 for (int i = 0; i < row; i++ ) { 15 for (int j = 0; j < col; j++) { 16 if (i == 0 && j == 0) { 17 continue; 18 } 19 if (i == 0) { 20 grid[i][j] += grid[i][j - 1]; // 前一列累加 21 } else if (j == 0) { 22 grid[i][j] += grid[i - 1][j]; 23 } else { 24 grid[i][j] += max(grid[i - 1][j], grid[i][j - 1]); 25 } 26 } 27 } 28 return grid[row - 1][col - 1]; 29 } 30 };
标签:row,Offer,int,47,grid,&&,size,dp,礼物 来源: https://www.cnblogs.com/MGFangel/p/16339370.html