June Leetcoding Challenge(6.21)

LeetCode六月挑战（6.21 ）Dungeon Game LeetCode 174解题方案

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.

Note:

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.

思路

该题使用动态规划解法。设骑士的初始血量为dp[0][0]=x，我们的目的就是求x，根据题意，骑士只可以往右走或者往下走，因此存在dp[i+1][j]和dp[i][j+1],如果是dungeon为负数，则dp[i+1][j]或dp[i][j+1]的值应为当前血量+下一步扣除的血量即dp[i+1][j] = dp[i][j] -dungeon[i+1][j]或dp[i][j+1] = dp[i][j] -dungeon[i][j+1]，直到判断到最后公主在的位置血量大于0即可。我们可以从公主所在位置往回推理求x,即将dp[i][j]设置在

dp[i][j] = max(1, min(dp[i + 1][j], dp[i][j + 1]) - dungeon[i][j]);

同时我们也需要考虑数组边界问题。

代码

java

class Solution {
    public int calculateMinimumHP(int[][] dungeon) {
        int i = dungeon.length-1;
        int j = dungeon[0].length-1;
        int[][] dp = new int[i+1][j+1];
        dp[i][j] = dungeon[i][j] > 0? 1: Math.abs(dungeon[i][j]) + 1;
        for(i=dungeon.length-1;i>=0;i--){
            for(j=dungeon[0].length-1;j>=0;j--){
                if (i == dungeon.length - 1 && j == dungeon[0].length - 1) {
                    continue;
                }
                dp[i][j] = Math.max(1, Math.min(i + 1 >= dungeon.length ? Integer.MAX_VALUE : dp[i + 1][j],
                    j + 1 >= dungeon[0].length ? Integer.MAX_VALUE : dp[i][j + 1]) - dungeon[i][j]);
            }
        }
        return dp[0][0];
    }
}