LeetCode---Triangle、House Robber、House Robber II
120. Triangle
Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.
For example, given the following triangle
[
[2],
[3,4],
[6,5,7],
[4,1,8,3]
]
The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11).
三角形最小路径和
给定一个三角形，找出自顶向下的最小路径和。每一步只能移动到下一行中相邻的结点上。
class Solution(object):
def minimumTotal(self, triangle):
"""
:type triangle: List[List[int]]
:rtype: int
"""
n=len(triangle)
dp=triangle[n-1]
for i in range(n-2,-1,-1):
for j in range(i+1):
dp[j]=min(dp[j],dp[j+1])+triangle[i][j]
return dp[0]198. House Robber
You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.
Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.
Example 1:
Input: [1,2,3,1] Output: 4 Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3). Total amount you can rob = 1 + 3 = 4.
Example 2:
Input: [2,7,9,3,1] Output: 12 Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1). Total amount you can rob = 2 + 9 + 1 = 12.
打家劫舍
你是一个专业的小偷，计划偷窃沿街的房屋。每间房内都藏有一定的现金，影响你偷窃的唯一制约因素就是相邻的房屋装有相互连通的防盗系统，如果两间相邻的房屋在同一晚上被小偷闯入，系统会自动报警。
给定一个代表每个房屋存放金额的非负整数数组，计算你在不触动警报装置的情况下，能够偷窃到的最高金额。
思路：考虑使用动态规划。dp[i]表示从0-i户可以打劫到的最大钱数。则有dp[i] = max(dp[i-1],dp[i-2]+nums[i])。
class Solution(object):
def rob(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if len(nums)==0:
return 0
if len(nums)==1:
return max(nums)
dp=[0]*len(nums)
dp[0]=nums[0]
dp[1]=max(nums[1],nums[0])
for i in range(2,len(nums)):
dp[i]=max(dp[i-1],dp[i-2]+nums[i])
return dp[len(nums)-1]213. House Robber II
You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.
Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.
Example 1:
Input: [2,3,2] Output: 3 Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2), because they are adjacent houses.
Example 2:
Input: [1,2,3,1] Output: 4 Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3). Total amount you can rob = 1 + 3 = 4.
你是一个专业的小偷，计划偷窃沿街的房屋，每间房内都藏有一定的现金。这个地方所有的房屋都围成一圈，这意味着第一个房屋和最后一个房屋是紧挨着的。同时，相邻的房屋装有相互连通的防盗系统，如果两间相邻的房屋在同一晚上被小偷闯入，系统会自动报警。
给定一个代表每个房屋存放金额的非负整数数组，计算你在不触动警报装置的情况下，能够偷窃到的最高金额。
思路：该题在上一个题的基础上让首位链接形成环，那么即表示第一个和最后一个不能同时被抢，则问题分解为House Robber（nums[0:len(nums)-1]）和House Robber（nums[1:len(nums)]），两者中比较大的那个即为结果。
class Solution(object):
def rob(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
length=len(nums)
if length==0:
return 0
elif length==1:
return nums[0]
elif length==2:
return max(nums[0],nums[1])
else:
return max(self.robrange(nums[:-1]),self.robrange(nums[1:]))
def robrange(self,nums):
if len(nums)==0:
return 0
if len(nums)==1:
return nums[0]
if len(nums)==2:
return max(nums[0],nums[1])
l=len(nums)
dp=[0]*l
dp[0]=nums[0]
dp[1]=max(nums[0],nums[1])
for i in range(2,len(nums)):
dp[i]=max(dp[i-2]+nums[i],dp[i-1])
return dp[len(nums)-1]