Dungeon Game

Dungeon Game

问题：

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.

思路：

　　动态规划

别人代码：

public class Solution {
    public int calculateMinimumHP(int[][] dungeon) {
        if(dungeon==null || dungeon.length==0 || dungeon[0].length==0)  return 0;
        int row = dungeon.length;
        int col = dungeon[0].length;
        
        dungeon[row-1][col-1] = dungeon[row-1][col-1]>=0 ? 0 : -dungeon[row-1][col-1];
        
        for(int j=col-2; j>=0; j--)
        {
            dungeon[row-1][j] = dungeon[row-1][j]>=dungeon[row-1][j+1] ? 0 : dungeon[row-1][j+1]-dungeon[row-1][j]; 
        }
        for(int i=row-2; i>=0; i--)
        {
            dungeon[i][col-1] = dungeon[i][col-1]>=dungeon[i+1][col-1] ? 0 : dungeon[i+1][col-1]-dungeon[i][col-1]; 
        }
        
        for(int i=row-2; i>=0; i--)
        {
            for(int j=col-2; j>=0; j--)
            {
                int down = dungeon[i][j]>=dungeon[i+1][j] ? 0 : dungeon[i+1][j]-dungeon[i][j];
                int right = dungeon[i][j]>=dungeon[i][j+1] ? 0 : dungeon[i][j+1]-dungeon[i][j]; 
                dungeon[i][j] = Math.min(down, right);
            }
        }
        return dungeon[0][0]+1;
    }
}

学习之处：

• 这道题一开始就想到用动态规划，也想到了要从右下角到左上角进行计算一步步的计算，但是就是没有写对动态规划方程，是参考同学的代码才写出来的，主要思路是，只看右侧来说，如果当前值>右边的值，那么不需要补充血量 直接为0，如果当前值>右边，则需要补充血量右边-当前值，关键之处，是首先要把右下角的值改成正值！！！！（我的代码未成功的根本原因之一）
• 亮点是从右上角到左上角进行遍历和计算，右下角值的处理是关键（可以把右下角的右侧和左侧值看成为0，行走那边不需要血量）
• 改变不好的习惯+1