【LeetCode】198. House Robber
198. House Robber
Description:
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.
解题思路：
图片来源：LeetCode 198. House Robber 递归->记忆->动态规划 ->新思路
Java语言实现：
解法一：递归法
时间复杂度和空间复杂度都太高，没有AC，直接Runtime Error，StackOverflowError
public class House_Robber {
public static void main(String[] args) {
// TODO Auto-generated method stub
int[] arr = {1, 2, 3, 1};
System.out.println(rob(arr));
}
public static int rob(int[] nums) {
return solve(nums, nums.length-1);
}
public static int solve(int[] nums, int n) {
if(n == 0) {
return nums[0];
}else if (n == 1) {
return Math.max(nums[0], nums[1]);
}else {
int A = solve(nums, n-2) + nums[n];
int B = solve(nums, n-1);
return Math.max(A, B);
}
}
}方法二：记忆化搜索方法
时间复杂度太高，没有AC，直接Time Limit Exceeded
import java.util.Arrays;
public class House_Robber {
public static void main(String[] args) {
// TODO Auto-generated method stub
int[] arr = {1, 2, 3, 1};
System.out.println(rob(arr));
}
public static int rob(int[] nums) {
if(nums.length == 0) {
return 0;
}else {
return solve1(nums, nums.length-1);
}
}
public static int solve1(int[] nums, int n) {
int[] memo = new int[n+1];
Arrays.fill(memo, -1);
if(n == 0) {
return nums[0];
}else if(n == 1) {
return Math.max(nums[0], nums[1]);
}else {
if(memo[n] == -1) {
int A = solve1(nums, n-2) + nums[n];
int B = solve1(nums, n-1);
memo[n] = Math.max(A, B);
}
}
return memo[n];
}
}方法三：动态规划方法
已经AC的代码
import java.util.Arrays;
public class House_Robber {
public static void main(String[] args) {
// TODO Auto-generated method stub
int[] arr = {1, 2, 3, 1};
System.out.println(rob(arr));
}
public static int rob(int[] nums) {
if(nums.length == 0) {
return 0;
}else {
return solve1(nums, nums.length-1);
}
}
//动态规划实现
public static int solve1(int[] nums, int n) {
int[] memo = new int[n+1];
Arrays.fill(memo, -1);
if(n == 0) {
return nums[0];
}else if (n == 1) {
return Math.max(nums[0], nums[1]);
}else {
memo[0] = nums[0];
memo[1] = Math.max(nums[0], nums[1]);
for(int j=2; j<=n; j++) {
if(memo[j] == -1) {
int A = memo[j-2] + nums[j];
int B = memo[j-1];
memo[j] = Math.max(A, B);
}
}
}
return memo[n];
}
}Python语言实现：
解法一：递归法
解法没有问题，就是Time Limit Exceeded。
class Solution:
# The first method : Recursion
def rob(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
return self.tryRob(nums, 0)
def tryRob(self, nums, index):
'''
:param nums: 考虑抢劫 nums[index:len(nums)]这个范围的所有房子
:param index:
:return:
'''
if(index >= len(nums)):
return 0
res = 0
for i in range(index,len(nums)):
res = max(res, nums[i] + self.tryRob(nums, i + 2))
return res
solution = Solution()
#test 1
#nums = [1, 2, 3, 1]
#test 2
nums = [2, 7, 9, 3, 1]
print(solution.rob(nums))
解法二：记忆化搜索法
已经AC的代码
class Solution:
# The second method : Memory search
def __init__(self):
# memo[i] 表示考虑抢劫nums[i:n]所能获得的最大收益
self.memo = []
def rob(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
self.memo = [-1 for _ in range(len(nums))]
return self.tryRob(nums, 0)
def tryRob(self, nums, index):
'''
:param nums:
:param index:
:return:
'''
if( index >= len(nums)):
return 0
if(self.memo[index] != -1):
return self.memo[index]
res = 0
for i in range(index, len(nums)):
res = max(res, nums[i] + self.tryRob(nums, i + 2))
self.memo[index] = res
return res
solution = Solution()
#test 1
#nums = [1, 2, 3, 1]
#test 2
nums = [2, 7, 9, 3, 1]
print(solution.rob(nums))解法三：动态规划法
已经AC的代码
class Solution:
def __init__(self):
# memo[i]表示考虑抢劫nums[i:n]所能获得的最大收益
self.memo = []
# The third method : dynamic programming
def rob(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
nums_len = len(nums)
if nums_len == 0:
return 0
self.memo = [-1 for _ in range(nums_len)]
self.memo[nums_len - 1] = nums[nums_len - 1]
for i in range(nums_len - 1, -1, -1):
for j in range(i, nums_len):
A = self.memo[i]
if j + 2 < nums_len :
B = nums[j] + self.memo[j + 2]
else:
B = nums[j]
self.memo[i] = max(A, B)
return self.memo[0]
solution = Solution()
#test 1
#nums = [1, 2, 3, 1]
#test 2
nums = [1, 2,]
print(solution.rob(nums))总结：
Java代码实现的动态规划和Python代码实现的动态规划，是两个不同的状态转移方程，这一点需要大家注意。
Reference：
https://blog.csdn.net/wys2011101169/article/details/73520653