LeetCode-213. House Robber II (JAVA)（有环）

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 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 tonightwithout alerting the police.

和House Robber题目类似，只是这次有了环形

加上条件的DP问题需要进行一定的转化才能成为标准的DP问题，这个转化就成了关键。

这里的关键就是最后那一个房间N和第一个房间相连了，可以这么进行转化：

考虑抢劫了第0个房间，那么问题就是求抢劫第0~N-1个房间的最大数。

考虑不抢劫第0个房间，那么问题就是求抢劫第1~N个房间的最大数。

public int rob(int[] nums) {if (nums == null)return 0;int n = nums.length;if (n == 0)return 0;if (n == 1)return nums[0];return Math.max(robDP(nums, 0, n - 2), robDP(nums, 1, n - 1));}int robDP(int[] nums, int first, int last) {int n = last - first + 1;if (n == 0)return 0;if (n == 1)return nums[first];int dp[] = new int[n];dp[0] = nums[first];// 注意下标dp[1] = Math.max(nums[first], nums[first + 1]);for (int i = 2; i < n; i++)dp[i] = Math.max(dp[i - 2] + nums[first + i], dp[i - 1]);return dp[n - 1];}

优化：

int rob(int[] nums, int first, int last) {// 因为只需要比较第dp[i-2]+nums[i]和dp[i-1]的大小，// 所以可以把House Robber I中的数组也省略掉，// 而只需要两个变量来存储着两个值，这样更能节约空间。// 这里pLast存储的是House Robber I中dp[i-1]的值，// ppLast存储的是House Robber I中dp[i-2]的值，//逻辑和House Robber I是一样的。int pLast = 0, ppLast = 0;for (int i = first; i <= last; i++) {int tmp = pLast;pLast = Math.max(ppLast + nums[i], pLast);ppLast = tmp;}return pLast;}

参考： LeetCode-198. House Robber (JAVA)寻找数组不相邻组合最大值DP