174. Dungeon Game

题目：

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 least 7 if he follows the optimal path RIGHT-> RIGHT -> DOWN -> DOWN.

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

Notes:

The knight's health has no upper bound.
Any room can contain threats or power-ups, even the first room the knight enters and the bottom-right room where the princess is imprisoned.

 

链接： http://leetcode.com/problems/dungeon-game/

题解：

题目好长啊...就是一个骑士要救公主，从迷宫坐上到右下，一共需要多少体力。 这里比较明显的想到了用DP，条件就是，当(i + 1, j) 和(i, j + 1)的值都小于0时，我们才更新(i，j)，否则我们不需要更多的体力。这样自右下角dp到左上角就可以了。 其实应该也可以用BFS或者其他的weighted shortest path方法。 也应该可以用滚动数组来进行dp，留给二刷吧。

Time Complexity - O(mn)， Space Complexity - O(mn)。  

public class Solution {
    public int calculateMinimumHP(int[][] dungeon) {
        if(dungeon == null || dungeon.length == 0)
            return 1;
        int m = dungeon.length, n = dungeon[0].length;
        
        for(int i = m - 2; i >= 0; i--) {             // initialize last row
            dungeon[i][n - 1] += Math.min(dungeon[i + 1][n - 1], 0);
        }   
            
        
        for(int j = n - 2; j >= 0; j--) {               // initialize last column
            dungeon[m - 1][j] += Math.min(dungeon[m - 1][j + 1], 0);
        }
        
        for(int i = dungeon.length - 2; i >= 0; i--) {          // dynamic programming
            for(int j = dungeon[0].length - 2; j >= 0; j--) {
                if(Math.max(dungeon[i + 1][j], dungeon[i][j + 1]) < 0)      // if right and down >= 0, we do not require more hp
                    dungeon[i][j] += Math.max(dungeon[i + 1][j], dungeon[i][j + 1]);
            }
        }
        
        return dungeon[0][0] >= 0 ? 1 : -dungeon[0][0] + 1;
    }
}

 

测试：