[LeetCode] 213. House Robber II 打家劫舍 II
Note: This is an extension of House Robber.
After robbing those houses on that street, the thief has found himself a new place for his thievery so that he will not get too much attention. This time, all houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, the security system for these houses remain the same as for those in the previous street.
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.
Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.
198. House Robber 的拓展，现在房子排成了一个圆圈，抢第一个屋子就不能抢最后一个屋子，抢最后一个屋子就不能抢第一个屋子。
解法：还是动态规划DP，分别算出这两个条件下的最大抢劫金额，然后取更大的就行。
Java:
public class Solution {
public int rob(int[] nums) {
return Math.max(rob(nums, 0), rob(nums, 1));
}
public int rob(int[] nums, int offset) {
// 如果长度过小，则直接返回结果
if(nums.length <= 1 + offset){
return nums.length <= offset ? 0 : nums[0 + offset];
}
int a = nums[0 + offset];
// 如果offset是1，则从下标为1的元素开始计算，所以要比较nums[1]和nums[2]
int b = Math.max(nums[0 + offset], nums[1 + offset]);
// 对于不抢劫最后一个房子的情况，i要小于nums.length - 1
for(int i = 2 + offset; i < nums.length - 1 + offset; i++){
int tmp = b;
b = Math.max(a + nums[i], b);
a = tmp;
}
return b;
}
}
Python:
class Solution:
# @param {integer[]} nums
# @return {integer}
def rob(self, nums):
if len(nums) == 0:
return 0
if len(nums) == 1:
return nums[0]
return max(self.robRange(nums, 0, len(nums) - 1),\
self.robRange(nums, 1, len(nums)))
def robRange(self, nums, start, end):
num_i, num_i_1 = nums[start], 0
for i in xrange(start + 1, end):
num_i_1, num_i_2 = num_i, num_i_1
num_i = max(nums[i] + num_i_2, num_i_1);
return num_i
C++:
class Solution {
public:
int rob(vector<int>& nums) {
if (nums.size() == 0) {
return 0;
}
if (nums.size() == 1) {
return nums[0];
}
return max(robRange(nums, 0, nums.size() - 1), // Include the first one of nums without the last one.
robRange(nums, 1, nums.size())); // Include the last one of nums without the first one.
}
int robRange(vector<int>& nums, int start, int end) {
int num_i = nums[start], num_i_1 = 0, num_i_2 = 0;
for (int i = start + 1; i < end; ++i) {
num_i_2 = num_i_1;
num_i_1 = num_i;
num_i = max(nums[i] + num_i_2, num_i_1);
}
return num_i;
}
};
C++:
class Solution {
public:
int rob(vector<int>& nums) {
if (nums.size() <= 1) return nums.empty() ? 0 : nums[0];
return max(rob(nums, 0, nums.size() - 1), rob(nums, 1, nums.size()));
}
int rob(vector<int> &nums, int left, int right) {
int a = 0, b = 0;
for (int i = left; i < right; ++i) {
int m = a, n = b;
a = n + nums[i];
b = max(m, n);
}
return max(a, b);
}
};
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[LeetCode] 198. House Robber 打家劫舍
