House Robber II——Leetcode

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.

题目大意：所有的住户围成一圈，只能隔着抢劫，问可以抢到的总额最大是多少。

解题思路：

因为第一家和最后一家是连着的，所以抢第一家就不能抢最后一家，抢最后一家就不能抢第一家，那么可以把数组分成两段，0~n-1和1~n两段，分别计算最大的，取较大的。

    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 res = -1;
        int[] max = new int[nums.length+1];
        max[0]=nums[0];
        max[1]=Math.max(nums[0],nums[1]);
        for(int i=2;i<nums.length-1;i++){
            max[i]=Math.max(max[i-1],max[i-2]+nums[i]);
        }
        res=Math.max(res,max[nums.length-2]);
        Arrays.fill(max,0);
        max[1]=nums[1];
        max[2]=Math.max(nums[1],nums[2]);
        for(int i=3;i<nums.length;i++){
            max[i]=Math.max(max[i-1],max[i-2]+nums[i]);
        }
        res=Math.max(res,max[nums.length-1]);
        return res;
    }