Apolo_Shane的博客

是198题. House Robber 的反例子# 1/小偷抢的越多，其剩下的就越少，其目标就是剩下的最少，但是不能连偷两个，这样的话会造成梯子有隔断。# 2/小偷不能偷相邻的，正好是梯子不能跳两格对应上class Solution:    def minCostClimbingStairs(self, cost):        last=now=0        f

#本题找规律容易l[i][j] = l[i - 1][j] + l[i][j - 1]但是一定要注意处理边界问题，通过生成（m+1)*(n+1)的二维数组来避免边界问题class Solution: def uniquePathsWithObstacles(self, obstacleGrid): m = len(obstacleGrid) n = l

#二维数组加上for循环class Solution: def uniquePaths(self, m, n): l=[[1]*n for i in range(m)]#生成m行n列的二维数组 for i in range(1,m): for j in range(1,n): l[i][j]=l[i

class Solution: def minPathSum(self, grid): m=len(grid) n=len(grid[0]) l=[[0]*(n) for i in range(m)] for i in range(m): for j in range(n):

#动态规划皆可用表格解决！#1\自上往下class Solution: def minimumTotal(self, triangle): m = len(triangle) l = triangle for i in range(1, m): for j in range(i + 1):

#典型的二分查找。 #注意边界问题。class Solution(object): def firstBadVersion(self, n): l=1 r=n while True: mid=(l+r)//2# if isBadVersion(mid):

class Solution(object): def numTrees(self, n): num = [[0] * (n + 1) for i in range(n + 1)] num[0][0] = 1 for i in range(1, n + 1): for j in range(1, n + 1):