Java for LeetCode 174 Dungeon Game

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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.

解题思路：

dp问题，有两种思路：

思路一：dp[i][j]表示从起点到dungeon[i][j]所需的最小血量，但是这种思路递推方程非常不好写

思路二：dp[i][j]表示从dungeon[i][j]到终点所需的最小血量，使用一维数组即可，递推方程

dp[j] = dungeon[i][j] >= Math.min(dp[j + 1], dp[j]) - 1 ? 1: Math.min(dp[j + 1], dp[j]) - dungeon[i][j]

当然，也可以直接用dungeon[i][j]数组替代dp[]数组，JAVA实现如下：

    public int calculateMinimumHP(int[][] dungeon) {
		int[] dp = new int[dungeon[0].length];
		dp[dungeon[0].length - 1] = dungeon[dungeon.length - 1][dungeon[0].length - 1] >= 0 ? 1
				: 1 - dungeon[dungeon.length - 1][dungeon[0].length - 1];
		for (int i = dungeon[0].length - 2; i >= 0; i--)
			dp[i] = dungeon[dungeon.length - 1][i] >= dp[i + 1] - 1 ? 1
					: dp[i + 1] - dungeon[dungeon.length - 1][i];
		for (int i = dungeon.length - 2; i >= 0; i--) {
			dp[dungeon[0].length - 1] = dungeon[i][dungeon[0].length - 1] >= dp[dungeon[0].length - 1] - 1 ? 1
					: dp[dungeon[0].length - 1]
							- dungeon[i][dungeon[0].length - 1];
			for (int j = dungeon[0].length - 2; j >= 0; j--)
				dp[j] = dungeon[i][j] >= Math.min(dp[j + 1], dp[j]) - 1 ? 1
						: Math.min(dp[j + 1], dp[j]) - dungeon[i][j];
		}
		return dp[0];
    }

Java for LeetCode 174 Dungeon Game

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原文地址：http://www.cnblogs.com/tonyluis/p/4557004.html