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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 arearranged 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.
此题思想是动态规划，声明两个变量，一个保存上一个被偷rob时最大值，另一个保存unrob上一个没有被偷的最大值，此时求下一个屋子是，rob=unrob+a[i]；unrob=上一次rob和unrob的最大值（因为不一定是偷和被偷时交叉进行的，中间可能有连续几个不偷）。本题是将房子连成环，所以需要访问两次，一次没有最后一个，从0到len-2;一次没有第一个，从1到length-1，最后求最大，此题需要遍历两遍。
public class Solution {
public int rob(int[] nums) {
if(nums.length==1)
return nums[0];
int rob=0;
int unrob=0;
for(int i=0;i<nums.length-1;i++){
int temp=unrob;
unrob=Math.max(rob,unrob);
rob=temp+nums[i];
}
int res1=Math.max(unrob,rob);
unrob=0;
rob=0;
for(int i=1;i<nums.length;i++){
int temp=unrob;
unrob=Math.max(rob,unrob);
rob=temp+nums[i];
}
int res2=Math.max(unrob,rob);
return Math.max(res1,res2);
}
}
还有一种写法，空间是O（n），一个数组表示第i个被偷，一个表示第i个没有被偷。代码如下。（非环）
public class Solution {
public int rob(int[] num) {
// 31 23 9 13 49 1 0
// 0 0 0 0
if(num==null || num.length==0) return 0;
int n = num.length;
int [] b = new int[n]; //include last element;
int [] d = new int[n]; //exclude last element;
b[0] = num[0];
d[0] = 0;
for(int i=1; i<n; i++) {
b[i] = d[i-1] + num[i];
d[i] = Math.max(b[i-1], d[i-1]);
}
return Math.max(d[n-1], b[n-1]);
}
}
维护一个数组，nums[i]表示到i时最大的钱数。
public class Solution {
//1 2 3
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]);
return Math.max(robsub(nums, 0, nums.length-2), robsub(nums, 1, nums.length-1));
}
private int robsub(int[] nums, int s, int e) {
int n = e - s + 1;
int[] d =new int[n];
d[0] = nums[s];
d[1] = Math.max(nums[s], nums[s+1]);
for(int i=2; i<n; i++) {
d[i] = Math.max(d[i-2]+nums[s+i], d[i-1]);
}
return d[n-1];
}
}
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原文地址：http://www.cnblogs.com/qiaomu/p/4619755.html
