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_213.java
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_213.java
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package com.fishercoder.solutions;
/**
* 213. House Robber 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.
*/
public class _213 {
public static class Solution1 {
/**
* Another dp problem:
* separate them into two cases:
* 1. rob from house 1 to n-1, get max1
* 2. rob from house 2 to n, get max2 take the max from the above two max
*/
public int rob(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
if (nums.length == 1) {
return nums[0];
}
if (nums.length == 2) {
return Math.max(nums[0], nums[1]);
}
int[] dp = new int[nums.length - 1];
//rob 1 to n-1
dp[0] = nums[0];
dp[1] = Math.max(nums[0], nums[1]);
for (int i = 2; i < nums.length - 1; i++) {
dp[i] = Math.max(dp[i - 2] + nums[i], dp[i - 1]);
}
int prevMax = dp[nums.length - 2];
//rob 2 to n
dp = new int[nums.length - 1];
dp[0] = nums[1];
dp[1] = Math.max(nums[1], nums[2]);
for (int i = 3; i < nums.length; i++) {
dp[i - 1] = Math.max(dp[i - 3] + nums[i], dp[i - 2]);
}
return Math.max(prevMax, dp[nums.length - 2]);
}
}
}