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，

8df1
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])