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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; }
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原文地址：http://www.cnblogs.com/aboutblank/p/4823725.html
