LeetCode-Escape The Ghosts

Description:
You are playing a simplified Pacman game. You start at the point (0, 0), and your destination is (target[0], target[1]). There are several ghosts on the map, the i-th ghost starts at (ghosts[i][0], ghosts[i][1]).

Each turn, you and all ghosts simultaneously may move in one of 4 cardinal directions: north, east, west, or south, going from the previous point to a new point 1 unit of distance away.

You escape if and only if you can reach the target before any ghost reaches you (for any given moves the ghosts may take.) If you reach any square (including the target) at the same time as a ghost, it doesn’t count as an escape.

Return True if and only if it is possible to escape.

Example 1:

Input: 
ghosts = [[1, 0], [0, 3]]
target = [0, 1]
Output: true
Explanation: 
You can directly reach the destination (0, 1) at time 1, while the ghosts located at (1, 0) or (0, 3) have no way to catch up with you.

Example 2:

Input: 
ghosts = [[1, 0]]
target = [2, 0]
Output: false
Explanation: 
You need to reach the destination (2, 0), but the ghost at (1, 0) lies between you and the destination.

Example 3:

Input: 
ghosts = [[2, 0]]
target = [1, 0]
Output: false
Explanation: 
The ghost can reach the target at the same time as you.

Note:

• All points have coordinates with absolute value <= 10000.
• The number of ghosts will not exceed 100.

题意：模拟一个吃豆豆的游戏，初始位置在(0, 0)点，需要到达目标位置(target[0], target[1])；在位置(ghosts[i][0], ghosts[i][1])上有幽灵；现在，需要计算我们是否可以在碰到幽灵之前到达目的地(需要注意一点的是，如果我们和幽灵同时达到目的地，游戏有相当于失败)；

解法：如果我们模拟自己的轨迹和幽灵的轨迹，那么计算量会非常的大，但是这道题目只需要我们判断是否可以在幽灵之前到达目的地，那么我们只需要判断我们到达目的地的最大距离是否比任何一个幽灵到达目的地的距离来的小即可；

Java

class Solution {
    public boolean escapeGhosts(int[][] ghosts, int[] target) {
        int maxDistance = Math.abs(target[0]) + Math.abs(target[1]);
        for (int[] ghost: ghosts) {
            if (Math.abs(ghost[0] - target[0]) + 
                Math.abs(ghost[1] - target[1]) <= maxDistance) {
                return false;
            }
        }
        return true;
    }
}