Leetcode 486 python 解题报告

AC代码：

class Solution(object):    def PRedictTheWinner(self, nums):        """        :type nums: List[int]        :rtype: bool        """        total = sum(nums)        sum1 = [[0 for i in range(len(nums))]for i in range(len(nums))]        dp = [[0 for i in range(len(nums)+1)] for i in range(len(nums)+1)]        for i in range(len(nums)):            dp[i][i] = nums[i]        for i in range(len(nums)):            for j in range(i,len(nums)):                sum1[i][j] = sum(nums[:j+1]) - sum(nums[:i])        print sum1        for k in range(1,len(nums)):            for i in range(len(nums)-k):                j = i + k                dp[i][j] = max(nums[i]+sum1[i+1][j] - dp[i+1][j],nums[j]+sum1[i][j-1]-dp[i][j-1])            #print dp        tmp = dp[0][len(nums)-1]        if tmp*2 >= total:            return True        else:            return False思路：定义两个二维数组，sum1和dp，sum1[i][j]表示的是i-j的总和，dp[i][j]则表示i-j中，先选择的人能获得的最大加和，
因为两人需要轮流选择，每人每次都会选择最有利于自己的结果，因此下次选择时，需从前面的总和减去前一组选择可获得的最大值，然后再比较先选左边还是先选右边能获得更大的结果，
于是可以得到动态转换方程：
dp[i][j] = max(nums[i]+sum1[i+1][j] - dp[i+1][j],nums[j]+sum1[i][j-1]-dp[i][j-1])