174 Dungeon Game

174. Dungeon Game

1. Question

The demons had captured the princess (P) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (K) was initially positioned in the top-left room and must fight his way through the dungeon to rescue the princess.

The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately.

Some of the rooms are guarded by demons, so the knight loses health (negative_integers) upon entering these rooms; other rooms are either empty (_0's) or contain magic orbs that increase the knight's health (_positive_integers).

In order to reach the princess as quickly as possible, the knight decides to move only rightward or downward in each step.

Write a function to determine the knight's minimum initial health so that he is able to rescue the princess.

For example, given the dungeon below, the initial health of the knight must be at least7if he follows the optimal pathRIGHT-> RIGHT -> DOWN -> DOWN.

-2(K)	-3	3
-5	-10	1
10	30	-5(P)

2. Implementation

(1) DP

思路: 这题和Unique Path的思路类似，不同在于现在问你在起始点的初始值最少可以是多少, 所以写法是从右上角到左上角，状态转移有4种情况:

(1) 当骑士在右下角时候，需要的血量至少是1，如果dungeon[m - 1][n - 1]是负数的话，则需要 1 - dungeon[m - 1][n - 1]，我们取两者中较大值，所以状态转移方程 dp[i][j] = Math.max(dp[i][j], 1 - dungeon[i][j])

(2) 当骑士在 m - 1行时，dp[i][j]的状态只取决于dp[i][j + 1], 状态转移方程式dp[i][j] = Math.max(1, dp[i][j + 1] - dungeon[i][j])

(3) 同理，当骑士在n - 1列时，dp[i][j] = Math.max(1, dp[i + 1][j] - dungeon[i][j])

(4) 当在骑士在其他行列时，dp[i][j]的状态取决于dp[i + 1][j]和dp[i][j + 1]之间的较小值, dp[i][j] = Math.max(1, Math.min(dp[i + 1][j], dp[i][j + 1]) - dungeon[i][j])

class Solution {
    public int calculateMinimumHP(int[][] dungeon) {
        if (dungeon == null || dungeon.length == 0) {
            return 0;
        }

        int m = dungeon.length, n = dungeon[0].length;
        int[][] dp = new int[m][n];

        for (int i = m - 1; i >= 0; i--) {
            for (int j = n - 1; j >= 0; j--) {
                if (i == m - 1 && j == n - 1) {
                    dp[i][j] = Math.max(1, 1 - dungeon[i][j]);
                }
                else if (i == m - 1) {
                    dp[i][j] = Math.max(1, dp[i][j + 1] - dungeon[i][j]);
                }
                else if (j == n - 1) {
                    dp[i][j] = Math.max(1, dp[i + 1][j] - dungeon[i][j]);
                }
                else {
                    dp[i][j] = Math.max(1, Math.min(dp[i + 1][j], dp[i][j + 1]) - dungeon[i][j]);
                }
            }
        }
        return dp[0][0];
    }
}

3. Time & Space Complexity

DP: 时间复杂度O(mn), 空间复杂度O(mn)