213. House-Robber-II

difficulty: Medium

section pre{ background-color: #eee; border: 1px solid #ddd; padding:10px; border-radius: 5px; }

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.

Method One

class Solution {
    public int rob(int[] nums) {
        if(nums == null || nums.length == 0) {
            return 0;
        }
        if( nums.length == 1 ) {
            return nums[0];
        }
        int max1 = simpleRob(nums, 0, nums.length - 1);
        int max2 = simpleRob(nums, 1, nums.length);
        return Math.max( max1, max2 );
    }
    public int simpleRob(int[] nums, int start, int end) {
        if( nums.length < 1 ) {
            return 0;
        }
        int prePre = 0;
        int pre = 0;
        for(int i = start; i < end; i++ ) {
            int temp = pre;
            pre = Math.max( prePre + nums[i], pre);
            prePre = temp;
        }
        return pre;
    }
}
​
/**
你不能同时抢首尾两个房子。
所以问题变成了，从 0 到 n-1 和 1 - n 两种抢法的最大收益的比较。
*/
​
​
​