URL： https://leetcode.com/problems/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.
解决方案：
Since it is a cyclied array, to make our dynamic programming alogrithm works, we need to break the cyclied array into two single-way arrays.
2-The priciple here is that either a particular house i-th can be robbed or not robbed at all.
For example 1 -> 3 -> 6 -> 5, we can break at the first position, so that the problem is divided to two parts,which are 1 -> 3 -> 6 (we choose to rob the first element, but not the last one.) and 3->6->5. Finally, we can reuse the House Robber I to solve these two sub-problems and pick out the optimal solution.
3-Here some of us may have the confusion that: we can rob house 1, it does not mean that we must rob house 1, whether we rob house 1 is depending on its next value. The correct complement of this statement is that: we should not(must not) rob house 1 ( and similarly, whether to rob the last house depends on the whole broken array.)
class Solution {
public int rob(int[] nums) {
if(nums == null || nums.length == 0) return 0;
if(nums.length == 1) return nums[0];
int len = nums.length;
int[] amount = new int[len];
//The first case: we can rob house 1, definitely, we cannot rob the last one
amount[0] = nums[0];
// whether we rob house 1 is depending the relative value
amount[1] = Math.max(nums[0],nums[1]);
for(int idx =2; idx < len-1; idx ++){
amount[idx] = Math.max(amount[idx-1],amount[idx-2] + nums[idx]);
}
// the last element is zero. amount[len-1] by default is zero
int totalSumCaseOne = amount[len-2];
// The second case: we don‘t rob house 1 at all.
amount = new int[len];
amount[0] = 0;
amount[1] = nums[1];
for(int idx = 2 ;idx< len; idx++){
amount[idx] = Math.max(amount[idx-1],amount[idx-2] + nums[idx]);
}
int totalSumCaseTwo = amount[len-1];
//select the largest one
return Math.max(totalSumCaseOne,totalSumCaseTwo);
}
}
原文地址：https://www.cnblogs.com/frankcui/p/10480695.html