LeetCode 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.


解题思路:这题和上一题的不同决定了这一题比上一题多了一次循环。上一题中,房子不是头尾相连的,而这一题里面,第一个房子和最后一个房子相连了,也就是说,强盗抢了第一个房子,就不能抢最后一个房子,那可以做两次循环,第一次循环保证抢了第一个房子,不抢最后一个房子,第二次循环保证抢了最后一个房子,不抢第一个房子,而中间的间隔抢劫规则是相同的,也就是指,我们所求的抢劫钱数的数组的对应关系和上一题一样。
public class Solution {
    public int rob(int[] nums) {
        if(nums.length<=0){
            return 0;
        }else if(nums.length<=3){
            int max=nums[0];
            for(int i=0;i<nums.length;i++){
                max=nums[i]>max?nums[i]:max;
            }
            return max;
        }else{
            int a[]= new int[nums.length]; //从第一个房子开始
            int b[]= new int[nums.length]; //从最后一个房子开始
            for(int i=0;i<nums.length;i++){ //初始化,可省略
                a[i]=0;
                b[i]=0;
            }
            
            a[0]=nums[0];
            a[1]=nums[1];
            a[2]=a[0]+nums[2];
            for(int i=3;i<nums.length-1;i++){
                a[i]=Math.max(a[i-2],a[i-3])+nums[i];
            }
            b[nums.length-1]=nums[nums.length-1];
            b[nums.length-2]=nums[nums.length-2];
            b[nums.length-3]=nums[nums.length-3]+b[nums.length-1];
            for(int i=nums.length-4;i>0;i--){
                b[i]=Math.max(b[i+2],b[i+3])+nums[i];
            }
            int max=a[0];
            for(int i=0;i<nums.length;i++){
                max = max>a[i]?max:a[i];
                max = max>b[i]?max:b[i];
            }
            return max;
        }
    }
}

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