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.
一刷
题解：其实就是比较自己离target的距离和其他所有ghost离target的最近距离，如果自己的距离比较小，则可以在ghost到达之前到达，因为对于ghost来说，在终点等是最划算的
class Solution(object):
def escapeGhosts(self, ghosts, target):
"""
:type ghosts: List[List[int]]
:type target: List[int]
:rtype: bool
"""
dist_my = abs(target[0]) + abs(target[1])
dist_ghost = sys.maxint
for ghost in ghosts:
dist_ghost = min(abs(target[0]- ghost[0]) + abs(target[1]- ghost[1]), dist_ghost)
return dist_my<=dist_ghost
