动态规划之理解「无后效性」

第 174 题：地下城游戏 官方题解 leetcode-cn.com/problems/du…

Java 代码：

import java.util.Arrays;

public class Solution {

    public int calculateMinimumHP(int[][] dungeon) {
        int rows = dungeon.length;
        int cols = dungeon[0].length;

        // 理解多加 1 位的原因
        int[][] dp = new int[rows + 1][cols + 1];
        for (int i = 0; i <= rows; i++) {
            Arrays.fill(dp[i], Integer.MAX_VALUE);
        }

        // 初始化，使得 dp[rows - 1][cols - 1] = 1
        dp[rows][cols - 1] = 1;
        dp[rows - 1][cols] = 1;

        for (int i = rows - 1; i >= 0; i--) {
            for (int j = cols - 1; j >= 0; j--) {
                int minVal = Math.min(dp[i + 1][j], dp[i][j + 1]);
                dp[i][j] = Math.max(minVal - dungeon[i][j], 1);
            }
        }
        return dp[0][0];
    }
}

Python 代码：

from typing import List


class Solution:
    def calculateMinimumHP(self, dungeon: List[List[int]]) -> int:
        rows = len(dungeon)
        cols = len(dungeon[0])

        dp = [[9999 for _ in range(cols + 1)] for _ in range(rows + 1)]

        dp[rows][cols - 1] = 1
        dp[rows - 1][cols] = 1
        for i in range(rows - 1, -1, -1):
            for j in range(cols - 1, -1, -1):
                min_val = min(dp[i + 1][j], dp[i][j + 1])
                dp[i][j] = max(min_val - dungeon[i][j], 1)
        return dp[0][0]

C++ 代码：

#include <iostream>
#include <vector>

using namespace std;

class Solution {
public:
    int calculateMinimumHP(vector<vector<int>> &dungeon) {
        int rows = dungeon.size();
        int cols = dungeon[0].size();
        vector<vector<int>> dp(rows + 1, vector<int>(cols + 1, INT_MAX));

        dp[rows][cols - 1] = 1;
        dp[rows - 1][cols] = 1;
        for (int i = rows - 1; i >= 0; i--) {
            for (int j = cols - 1; j >= 0; j--) {
                int minVal = min(dp[i + 1][j], dp[i][j + 1]);
                dp[i][j] = max(minVal - dungeon[i][j], 1);
            }
        }
        return dp[0][0];
    }
};