## 题目描述：

LeetCode 488. Zuma Game

Think about Zuma Game. You have a row of balls on the table, colored red(R), yellow(Y), blue(B), green(G), and white(W). You also have several balls in your hand.

Each time, you may choose a ball in your hand, and insert it into the row (including the leftmost place and rightmost place). Then, if there is a group of 3 or more balls in the same color touching, remove these balls. Keep doing this until no more balls can be removed.

Find the minimal balls you have to insert to remove all the balls on the table. If you cannot remove all the balls, output -1.

```Examples:

Input: "WRRBBW", "RB" Output: -1 Explanation: WRRBBW -> WRR[R]BBW -> WBBW -> WBB[B]W -> WW Input: "WWRRBBWW", "WRBRW" Output: 2 Explanation: WWRRBBWW -> WWRR[R]BBWW -> WWBBWW -> WWBB[B]WW -> WWWW -> empty Input:"G", "GGGGG" Output: 2 Explanation: G -> G[G] -> GG[G] -> empty Input: "RBYYBBRRB", "YRBGB" Output: 3 Explanation: RBYYBBRRB -> RBYY[Y]BBRRB -> RBBBRRB -> RRRB -> B -> B[B] -> BB[B] -> empty
```

Note:

1. You may assume that the initial row of balls on the table won’t have any 3 or more consecutive balls with the same color.
2. The number of balls on the table won't exceed 20, and the string represents these balls is called "board" in the input.
3. The number of balls in your hand won't exceed 5, and the string represents these balls is called "hand" in the input.
4. Both input strings will be non-empty and only contain characters 'R','Y','B','G','W'.

## 题目大意：

1. 你可以假设初始桌子上不包含3个或者3个以上连续相同颜色的小球。
2. 桌子上小球的数目不会超过20，输入字符串记为board。
3. 手中的小球数目不会超过5，输入字符串记为hand。
4. 输入字符串非空，并且只包含'R','Y','B','G','W'。

## 解题思路：

```1. 每一次插入小球（1个或者2个相同颜色球）至少引发一次“消除动作”，否则就不插入

2. 对于board中出现的连续相同颜色，只在最右侧进行插入```

## Python代码：

``````class Solution(object):
def __init__(self):
self.cache = dict()

def findMinStep(self, board, hand):
"""
:type board: str
:type hand: str
:rtype: int
"""
if not board: return 0
if not hand: return -1
if board in self.cache: return self.cache[board]
cnt = collections.Counter(hand)
size = len(board)
ans = -1
num = 0
for x in range(1, len(board) + 1):
if x < len(board) and board[x] == board[x - 1]:
num += 1
else:
if cnt[board[x - 1]] + num > 1:
step = max(0, 2 - num)
cnt[board[x - 1]] -= step
nhand = ''.join(k * v for k, v in cnt.iteritems())
nboard = self.simplify(board[:x - num -1] + board[x:])
nans = self.findMinStep(nboard, nhand)
cnt[board[x - 1]] += step
if nans != -1 and (ans == -1 or nans + step < ans):
ans = nans + step
num = 0
self.cache[board] = ans
return ans

def simplify(self, board):
ptn = r'R{3,}|Y{3,}|B{3,}|G{3,}|W{3,}'
while re.search(ptn, board):
board = re.sub(ptn, '', board)
return board
``````

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