Dungeon Game

Dungeon Game

問題：

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：

　　動態規劃

別人代碼：

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;
    }
}

View Code

學習之處：

• 這道題一開始就想到用動態規劃，也想到了要從右下角到左上角進行計算一步步的計算，可是就是沒有寫對動態規劃方程，是參考同窗的代碼才寫出來的，主要思路是，只看右側來講，若是當前值>右邊的值，那麼不須要補充血量 直接爲0，若是當前值>右邊，則須要補充血量右邊-當前值，關鍵之處，是首先要把右下角的值改爲正值！！！！（個人代碼未成功的根本緣由之一）
• 亮點是從右上角到左上角進行遍歷和計算，右下角值的處理是關鍵（能夠把右下角的右側和左側值當作爲0，行走那邊不須要血量）
• 改變很差的習慣+1