dp = [ [0 for j in range(i+1)] for i in range(n)] dp[0][0] = value(1,1) for i in range(1, n): j = 0 while j <= i: # 如果是第一列,等于自身+上一位置的状态值 if j == 0: dp[i][j] = value(i+1,j+1) + dp[i - 1][j] # 否则,等于自身+(上一位置和左...
#在i之前的某个位置进行修改,得到的最大连续子数组和是 dp[i-1][1] + nums[i] dp[i][1] = max(dp[i-1][0]+x, dp[i-1][1]+nums[i]) # 更新答案 ans = max(ans, max(dp[i])) returnans t = int(input) # 遍历t次,表示t次询问 for_inrange(t): # 获得本次询问的数组长度n和...
cache = [[None] * nfor_inrange(m)]returnself.process(text1, text2,0,0, cache)defprocess(self, s1, s2, i, j, cache):# 当达到其中一个字符串的末尾时,递归结束iflen(s1) == iorlen(s2) == j:return0# 当cache[i][j]为None的时候,说明还没有计算过,计算后再返回ifcache[i][j]isN...
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n,m=[int(i) for i in input().split()] dp=[0 for j in range(n)] sum1=0 dp[0]=1 for j in range(1,m): dp[j]=2**(j) #print(dp) for i in range(n): _牛客网_牛客在手,offer不愁
dp[0]=0foriinrange(mx):# 如果 dp[i] 为 -1 ,则使用状态 i 中的火柴棍,# 无法拼出满足题意的正方形的边,直接处理下一个状态ifdp[i]==-1:continue# 枚举接下来要使用的火柴棍forjinrange(n):# 如果第 j 根火柴棍不在状态 i 中,且加入后拼出的边长不超过 target ,# 则状态 i | (1 << ...
时间复杂度为n,空间复杂度为n class Solution(object): def longestValidParentheses(self, s): """ :type s: str :rtype: int """ f=[0]*(len(s)+1) max_len = 0 for i in range(1,len(s)): if s[i] == ')': if s[i-1]== '(': ...
we have an array ofnnelements, for simplicity, numbers. To sort the subarray[l...r][l...r], we sort the subarrays[l...m][l...m]and[m+1...r][m+1...r], wherel≤m≤rl≤m≤r, then merge the two sorted arrays into one. Let's see how to do this in the easiest way...
classSolution(object):defuniquePaths(self, m, n):""" :type m: int :type n: int :rtype: int """dp = [[0forjinrange(n)]foriinrange(m)]foriinrange(m): dp[i][0] =1forjinrange(n): dp[0][j] =1foriinrange(1, m):forjinrange(1, n): ...
dp=[[0 for i in range(0,m+1)] for i in range(0,n+1)] for i in range(0,n): v[i],w[i]=map(int,input().split()) for i in range(0,n): for j in range(0,m+1): dp[i][j]=dp[i-1][j] if j>=v[i]: