【leetcode】1046. Last Stone Weight

题目如下：

We have a collection of rocks, each rock has a positive integer weight.

Each turn, we choose the two heaviest rocks and smash them together.  Suppose the stones have weights x and y with x <= y.  The result of this smash is:

• If x == y, both stones are totally destroyed;
• If x != y, the stone of weight x is totally destroyed, and the stone of weight y has new weight y-x.

At the end, there is at most 1 stone left.  Return the weight of this stone (or 0 if there are no stones left.)

Example 1:

Input: [2,7,4,1,8,1]
Output: 1
Explanation: 
We combine 7 and 8 to get 1 so the array converts to [2,4,1,1,1] then,
we combine 2 and 4 to get 2 so the array converts to [2,1,1,1] then,
we combine 2 and 1 to get 1 so the array converts to [1,1,1] then,
we combine 1 and 1 to get 0 so the array converts to [1] then that's the value of last stone.

Note:

1. 1 <= stones.length <= 30
2. 1 <= stones[i] <= 1000

解题思路：非常简单的题目，每次取最大的两个数，如果存在多个最大的两个数，随机取其中两个即可。

代码如下：

class Solution(object):
    def lastStoneWeight(self, stones):
        """
        :type stones: List[int]
        :rtype: int
        """
        while len(stones) > 1:
            max_index = 0
            second_max_index = 0
            ol = sorted(stones)
            for i in range(len(stones)):
                if ol[-1] == stones[i]:
                    max_index = i
                elif ol[-2] == stones[i] and max_index != i:
                    second_max_index = i
            if stones[max_index] == stones[second_max_index]:
                if max_index > second_max_index:
                    del stones[max_index]
                    del stones[second_max_index]
            else:
                stones[max_index] -= stones[second_max_index]
                del stones[second_max_index]
            #print stones
        return 0 if len(stones) == 0 else stones[0]

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